При решении уравнения для геометрии в сферических координатах возникает следующее уравнение для полярного угла
где Θ ( z ) \Theta(z) Θ ( z ) m m m γ 1 2 \gamma_1^2 γ 1 2
( 1 − z 2 ) ⋅ d 2 Θ ( z ) d z 2 − 2 ⋅ z ⋅ d Θ ( z ) d z + [ γ 1 2 − m 2 1 − z 2 ] ⋅ Θ ( z ) = 0. (1 - z^2) \cdot \frac{\displaystyle d^2 \Theta(z)}{\displaystyle dz^2} - 2 \cdot z \cdot \frac{\displaystyle d \Theta(z)}{\displaystyle dz} + \left[ \gamma_1^2 - \frac{\displaystyle m^2}{\displaystyle 1 - z^2} \right] \cdot \Theta(z) = 0. ( 1 − z 2 ) ⋅ d z 2 d 2 Θ ( z ) − 2 ⋅ z ⋅ d z d Θ ( z ) + [ γ 1 2 − 1 − z 2 m 2 ] ⋅ Θ ( z ) = 0. Заметим, что при m = 0 m = 0 m = 0
( 1 − z 2 ) ⋅ d 2 Θ ( z ) d z 2 − 2 ⋅ z ⋅ d Θ ( z ) d z + γ 1 2 ⋅ Θ ( z ) = 0 , − 1 < z < 1. (1 - z^2) \cdot \frac{\displaystyle d^2 \Theta(z)}{\displaystyle dz^2} - 2 \cdot z \cdot \frac{\displaystyle d \Theta(z)}{\displaystyle dz} + \gamma_1^2 \cdot \Theta(z) = 0, \quad -1 < z < 1. ( 1 − z 2 ) ⋅ d z 2 d 2 Θ ( z ) − 2 ⋅ z ⋅ d z d Θ ( z ) + γ 1 2 ⋅ Θ ( z ) = 0 , − 1 < z < 1. Это уравнение имеет ограниченные на интервале − 1 < z < 1 -1 < z < 1 − 1 < z < 1 γ 1 k 2 = k ⋅ ( k + 1 ) \gamma_{1k}^2 = k \cdot (k + 1) γ 1 k 2 = k ⋅ ( k + 1 ) k ∈ ( 0.. ∞ ) k \in (0..\infty) k ∈ ( 0..∞ )
где P k ( z ) P_k(z) P k ( z ) E.1 d d z ( ( 1 − z 2 ) ⋅ d Θ ( z ) d z ) + [ γ 1 2 − m 2 1 − z 2 ] ⋅ Θ ( z ) = 0 , − 1 < z < 1 , \frac{\displaystyle d}{\displaystyle dz} \left( (1 - z^2) \cdot \frac{\displaystyle d \Theta(z)}{\displaystyle dz} \right) + \left[ \gamma_1^2 - \frac{\displaystyle m^2}{\displaystyle 1 - z^2} \right] \cdot \Theta(z) = 0, \quad -1 < z < 1, d z d ( ( 1 − z 2 ) ⋅ d z d Θ ( z ) ) + [ γ 1 2 − 1 − z 2 m 2 ] ⋅ Θ ( z ) = 0 , − 1 < z < 1 , γ 1 k 2 = k ⋅ ( k + 1 ) \gamma_{1k}^2 = k \cdot (k + 1) γ 1 k 2 = k ⋅ ( k + 1 )
Сделаем подстановку Θ k m ( z ) = ( 1 − z 2 ) m 2 ⋅ Θ ^ k m ( z ) \Theta_{km}(z) = \left( 1 - z^2 \right)^{\frac{\displaystyle m}{\displaystyle 2}} \cdot \widehat{\Theta}_{km}(z) Θ k m ( z ) = ( 1 − z 2 ) 2 m ⋅ Θ k m ( z ) E.3 ( 1 − z 2 ) ⋅ d 2 Θ k m ( z ) d z 2 − 2 ⋅ z ⋅ d Θ k m ( z ) d z + [ k ⋅ ( k + 1 ) − m 2 1 − z 2 ] ⋅ Θ k m ( z ) = 0 , k ∈ ( 0.. ∞ ) . (1 - z^2) \cdot \frac{\displaystyle d^2 \Theta_{km}(z)}{\displaystyle dz^2} - 2 \cdot z \cdot \frac{\displaystyle d \Theta_{km}(z)}{\displaystyle dz} + \left[ k \cdot (k + 1) - \frac{\displaystyle m^2}{\displaystyle 1 - z^2} \right] \cdot \Theta_{km}(z) = 0, \quad k \in (0..\infty). ( 1 − z 2 ) ⋅ d z 2 d 2 Θ k m ( z ) − 2 ⋅ z ⋅ d z d Θ k m ( z ) + [ k ⋅ ( k + 1 ) − 1 − z 2 m 2 ] ⋅ Θ k m ( z ) = 0 , k ∈ ( 0..∞ ) .
d d z [ ( 1 − z 2 ) m 2 ⋅ Θ ^ k m ( z ) ] = ( 1 − z 2 ) m 2 ⋅ d Θ ^ k m ( z ) d z − m ⋅ z ⋅ ( 1 − z 2 ) m 2 − 1 ⋅ Θ ^ k m ( z ) . \frac{\displaystyle d}{\displaystyle dz} \left[ (1 - z^2)^{\frac{\displaystyle m}{\displaystyle 2}} \cdot \widehat{\Theta}_{km}(z) \right] = (1 - z^2)^{\frac{\displaystyle m}{\displaystyle 2}} \cdot \frac{\displaystyle d \widehat{\Theta}_{km}(z)}{\displaystyle dz} - m \cdot z \cdot (1 - z^2)^{\frac{\displaystyle m}{\displaystyle 2} - 1} \cdot \widehat{\Theta}_{km}(z). d z d [ ( 1 − z 2 ) 2 m ⋅ Θ k m ( z ) ] = ( 1 − z 2 ) 2 m ⋅ d z d Θ k m ( z ) − m ⋅ z ⋅ ( 1 − z 2 ) 2 m − 1 ⋅ Θ k m ( z ) . Для вычисления второй производной предварительно вычислим
d d z [ z 1 − z 2 ] = 1 1 − z 2 + 2 ⋅ z 2 ( 1 − z 2 ) 2 = 1 + z 2 ( 1 − z 2 ) 2 . \frac{\displaystyle d}{\displaystyle dz} \left[ \frac{\displaystyle z}{\displaystyle 1 - z^2} \right] = \frac{\displaystyle 1}{\displaystyle 1 - z^2} + \frac{\displaystyle 2 \cdot z^2}{\displaystyle (1 - z^2)^2} = \frac{\displaystyle 1 + z^2}{\displaystyle (1 - z^2)^2}. d z d [ 1 − z 2 z ] = 1 − z 2 1 + ( 1 − z 2 ) 2 2 ⋅ z 2 = ( 1 − z 2 ) 2 1 + z 2 . Тогда вторая производная запишется в виде
d 2 d z 2 [ ( 1 − z 2 ) m 2 ⋅ Θ ^ k m ( z ) ] = ( 1 − z 2 ) m 2 ⋅ d 2 Θ ^ k m ( z ) d z 2 − m ⋅ z ⋅ ( 1 − z 2 ) m 2 − 1 ⋅ d Θ ^ k m ( z ) d z − m ⋅ z 1 − z 2 ⋅ [ ( 1 − z 2 ) m 2 ⋅ d Θ ^ k m ( z ) d z − m ⋅ z ⋅ ( 1 − z 2 ) m 2 − 1 ⋅ Θ ^ k m ( z ) ] − m ⋅ 1 + z 2 ( 1 − z 2 ) 2 ⋅ ( 1 − z 2 ) m 2 ⋅ Θ ^ k m ( z ) . \begin{split}
&\frac{\displaystyle d^2}{\displaystyle dz^2} \left[ (1 - z^2)^{\frac{\displaystyle m}{\displaystyle 2}} \cdot \widehat{\Theta}_{km}(z) \right] = (1 - z^2)^{\frac{\displaystyle m}{\displaystyle 2}} \cdot \frac{\displaystyle d^2 \widehat{\Theta}_{km}(z)}{\displaystyle dz^2} - m \cdot z \cdot (1 - z^2)^{\frac{\displaystyle m}{\displaystyle 2} - 1} \cdot \frac{\displaystyle d \widehat{\Theta}_{km}(z)}{\displaystyle dz} - \\
&m \cdot \frac{\displaystyle z}{\displaystyle 1 - z^2} \cdot \left[ (1 - z^2)^{\frac{\displaystyle m}{\displaystyle 2}} \cdot \frac{\displaystyle d \widehat{\Theta}_{km}(z)}{\displaystyle dz} - m \cdot z \cdot (1 - z^2)^{\frac{\displaystyle m}{\displaystyle 2} - 1} \cdot \widehat{\Theta}_{km}(z) \right] -\\
&m \cdot \frac{\displaystyle 1 + z^2}{\displaystyle (1 - z^2)^2} \cdot (1 - z^2)^{\frac{\displaystyle m}{\displaystyle 2}} \cdot \widehat{\Theta}_{km}(z).
\end{split} d z 2 d 2 [ ( 1 − z 2 ) 2 m ⋅ Θ k m ( z ) ] = ( 1 − z 2 ) 2 m ⋅ d z 2 d 2 Θ k m ( z ) − m ⋅ z ⋅ ( 1 − z 2 ) 2 m − 1 ⋅ d z d Θ k m ( z ) − m ⋅ 1 − z 2 z ⋅ [ ( 1 − z 2 ) 2 m ⋅ d z d Θ k m ( z ) − m ⋅ z ⋅ ( 1 − z 2 ) 2 m − 1 ⋅ Θ k m ( z ) ] − m ⋅ ( 1 − z 2 ) 2 1 + z 2 ⋅ ( 1 − z 2 ) 2 m ⋅ Θ k m ( z ) . После приведения подобных слагаемых она примет вид
d 2 d z 2 [ ( 1 − z 2 ) m 2 ⋅ Θ ^ k m ( z ) ] = ( 1 − z 2 ) m 2 ⋅ d 2 Θ ^ k m ( z ) d z 2 − 2 ⋅ m ⋅ z 1 − z 2 ⋅ ( 1 − z 2 ) m 2 ⋅ d Θ ^ k m ( z ) d z + m ⋅ m ⋅ z 2 − 1 − z 2 ( 1 − z 2 ) 2 ⋅ ( 1 − z 2 ) m 2 ⋅ Θ ^ k m ( z ) . \begin{split}
&\frac{\displaystyle d^2}{\displaystyle dz^2} \left[ (1 - z^2)^{\frac{\displaystyle m}{\displaystyle 2}} \cdot \widehat{\Theta}_{km}(z) \right] = (1 - z^2)^{\frac{\displaystyle m}{\displaystyle 2}} \cdot \frac{\displaystyle d^2 \widehat{\Theta}_{km}(z)}{\displaystyle dz^2} - \frac{\displaystyle 2 \cdot m \cdot z}{\displaystyle 1 - z^2} \cdot (1 - z^2)^{\frac{\displaystyle m}{\displaystyle 2}} \cdot \frac{\displaystyle d \widehat{\Theta}_{km}(z)}{\displaystyle dz} +\\
&m \cdot \frac{\displaystyle m \cdot z^2 - 1 - z^2}{\displaystyle (1 - z^2)^2} \cdot (1 - z^2)^{\frac{\displaystyle m}{\displaystyle 2}} \cdot \widehat{\Theta}_{km}(z).
\end{split} d z 2 d 2 [ ( 1 − z 2 ) 2 m ⋅ Θ k m ( z ) ] = ( 1 − z 2 ) 2 m ⋅ d z 2 d 2 Θ k m ( z ) − 1 − z 2 2 ⋅ m ⋅ z ⋅ ( 1 − z 2 ) 2 m ⋅ d z d Θ k m ( z ) + m ⋅ ( 1 − z 2 ) 2 m ⋅ z 2 − 1 − z 2 ⋅ ( 1 − z 2 ) 2 m ⋅ Θ k m ( z ) . Подставим всё в уравнение (E.3 ( 1 − z 2 ) ⋅ d 2 Θ k m ( z ) d z 2 − 2 ⋅ z ⋅ d Θ k m ( z ) d z + [ k ⋅ ( k + 1 ) − m 2 1 − z 2 ] ⋅ Θ k m ( z ) = 0 , k ∈ ( 0.. ∞ ) . (1 - z^2) \cdot \frac{\displaystyle d^2 \Theta_{km}(z)}{\displaystyle dz^2} - 2 \cdot z \cdot \frac{\displaystyle d \Theta_{km}(z)}{\displaystyle dz} + \left[ k \cdot (k + 1) - \frac{\displaystyle m^2}{\displaystyle 1 - z^2} \right] \cdot \Theta_{km}(z) = 0, \quad k \in (0..\infty). ( 1 − z 2 ) ⋅ d z 2 d 2 Θ k m ( z ) − 2 ⋅ z ⋅ d z d Θ k m ( z ) + [ k ⋅ ( k + 1 ) − 1 − z 2 m 2 ] ⋅ Θ k m ( z ) = 0 , k ∈ ( 0..∞ ) . ( 1 − z 2 ) m 2 (1 - z^2)^{\frac{\displaystyle m}{\displaystyle 2}} ( 1 − z 2 ) 2 m
( 1 − z 2 ) ⋅ [ d 2 Θ ^ k m ( z ) d z 2 − 2 ⋅ m ⋅ z 1 − z 2 ⋅ d Θ ^ k m ( z ) d z + m ⋅ m ⋅ z 2 − 1 − z 2 ( 1 − z 2 ) 2 ⋅ Θ ^ k m ( z ) ] − 2 ⋅ z ⋅ [ d Θ ^ k m ( z ) d z − m ⋅ z 1 − z 2 ⋅ Θ ^ k m ( z ) ] + [ k ⋅ ( k + 1 ) − m 2 1 − z 2 ] ⋅ Θ ^ k m ( z ) = 0. \begin{split}
&(1 - z^2) \cdot \left[ \frac{\displaystyle d^2 \widehat{\Theta}_{km}(z)}{\displaystyle dz^2} - \frac{\displaystyle 2 \cdot m \cdot z}{\displaystyle 1 - z^2} \cdot \frac{\displaystyle d \widehat{\Theta}_{km}(z)}{\displaystyle dz} + m \cdot \frac{\displaystyle m \cdot z^2 - 1 - z^2}{\displaystyle (1 - z^2)^2} \cdot \widehat{\Theta}_{km}(z) \right] -\\
&2 \cdot z \cdot \left[ \frac{\displaystyle d \widehat{\Theta}_{km}(z)}{\displaystyle dz} - \frac{\displaystyle m \cdot z}{\displaystyle 1 - z^2} \cdot \widehat{\Theta}_{km}(z) \right] + \left[ k \cdot (k + 1) - \frac{\displaystyle m^2}{\displaystyle 1 - z^2} \right] \cdot \widehat{\Theta}_{km}(z) = 0.
\end{split} ( 1 − z 2 ) ⋅ [ d z 2 d 2 Θ k m ( z ) − 1 − z 2 2 ⋅ m ⋅ z ⋅ d z d Θ k m ( z ) + m ⋅ ( 1 − z 2 ) 2 m ⋅ z 2 − 1 − z 2 ⋅ Θ k m ( z ) ] − 2 ⋅ z ⋅ [ d z d Θ k m ( z ) − 1 − z 2 m ⋅ z ⋅ Θ k m ( z ) ] + [ k ⋅ ( k + 1 ) − 1 − z 2 m 2 ] ⋅ Θ k m ( z ) = 0. Сгруппируем коэффициенты при производных:
ζ 1 ⋅ d 2 Θ ^ k m ( z ) d z 2 + ζ 2 ⋅ d Θ ^ k m ( z ) d z + ζ 3 ⋅ Θ ^ k m ( z ) = 0 , \zeta_1 \cdot \frac{\displaystyle d^2 \widehat{\Theta}_{km}(z)}{\displaystyle dz^2} + \zeta_2 \cdot \frac{\displaystyle d \widehat{\Theta}_{km}(z)}{\displaystyle dz} + \zeta_3 \cdot \widehat{\Theta}_{km}(z) = 0, ζ 1 ⋅ d z 2 d 2 Θ k m ( z ) + ζ 2 ⋅ d z d Θ k m ( z ) + ζ 3 ⋅ Θ k m ( z ) = 0 , ζ 1 = 1 − z 2 , \zeta_1 = 1 - z^2, ζ 1 = 1 − z 2 , ζ 2 = − 2 ⋅ m ⋅ z − 2 ⋅ z = − 2 ⋅ ( m + 1 ) ⋅ z , \zeta_2 = -2 \cdot m \cdot z - 2 \cdot z = -2 \cdot (m + 1) \cdot z, ζ 2 = − 2 ⋅ m ⋅ z − 2 ⋅ z = − 2 ⋅ ( m + 1 ) ⋅ z , ζ 3 = m ⋅ m ⋅ z 2 − 1 − z 2 1 − z 2 + 2 ⋅ m ⋅ z 2 1 − z 2 + k ⋅ ( k + 1 ) − m 2 1 − z 2 = m 2 ⋅ z 2 − m − m ⋅ z 2 + 2 ⋅ m ⋅ z 2 − m 2 1 − z 2 + k ⋅ ( k + 1 ) = − m 2 ⋅ ( 1 − z 2 ) + m ⋅ ( 1 − z 2 ) 1 − z 2 + k ⋅ ( k + 1 ) = − m 2 − m + k ⋅ ( k + 1 ) = ( k − m ) ⋅ ( k + m + 1 ) , \begin{split}
&\zeta_3 = m \cdot \frac{\displaystyle m \cdot z^2 - 1 - z^2}{\displaystyle 1 - z^2} + \frac{\displaystyle 2 \cdot m \cdot z^2}{\displaystyle 1 - z^2} + k \cdot (k + 1) - \frac{\displaystyle m^2}{\displaystyle 1 - z^2} =\\
&\frac{\displaystyle m^2 \cdot z^2 - m - m \cdot z^2 + 2 \cdot m \cdot z^2 - m^2}{\displaystyle 1 - z^2} + k \cdot (k + 1) = - \frac{\displaystyle m^2 \cdot (1 - z^2) + m \cdot (1 - z^2)}{\displaystyle 1 - z^2} +\\
&k \cdot (k + 1) = - m^2 - m + k \cdot (k + 1) = (k - m) \cdot (k + m + 1),
\end{split} ζ 3 = m ⋅ 1 − z 2 m ⋅ z 2 − 1 − z 2 + 1 − z 2 2 ⋅ m ⋅ z 2 + k ⋅ ( k + 1 ) − 1 − z 2 m 2 = 1 − z 2 m 2 ⋅ z 2 − m − m ⋅ z 2 + 2 ⋅ m ⋅ z 2 − m 2 + k ⋅ ( k + 1 ) = − 1 − z 2 m 2 ⋅ ( 1 − z 2 ) + m ⋅ ( 1 − z 2 ) + k ⋅ ( k + 1 ) = − m 2 − m + k ⋅ ( k + 1 ) = ( k − m ) ⋅ ( k + m + 1 ) , таким образом, после подстановки уравнение (E.3 ( 1 − z 2 ) ⋅ d 2 Θ k m ( z ) d z 2 − 2 ⋅ z ⋅ d Θ k m ( z ) d z + [ k ⋅ ( k + 1 ) − m 2 1 − z 2 ] ⋅ Θ k m ( z ) = 0 , k ∈ ( 0.. ∞ ) . (1 - z^2) \cdot \frac{\displaystyle d^2 \Theta_{km}(z)}{\displaystyle dz^2} - 2 \cdot z \cdot \frac{\displaystyle d \Theta_{km}(z)}{\displaystyle dz} + \left[ k \cdot (k + 1) - \frac{\displaystyle m^2}{\displaystyle 1 - z^2} \right] \cdot \Theta_{km}(z) = 0, \quad k \in (0..\infty). ( 1 − z 2 ) ⋅ d z 2 d 2 Θ k m ( z ) − 2 ⋅ z ⋅ d z d Θ k m ( z ) + [ k ⋅ ( k + 1 ) − 1 − z 2 m 2 ] ⋅ Θ k m ( z ) = 0 , k ∈ ( 0..∞ ) .
Возьмём уравнение (E.3 ( 1 − z 2 ) ⋅ d 2 Θ k m ( z ) d z 2 − 2 ⋅ z ⋅ d Θ k m ( z ) d z + [ k ⋅ ( k + 1 ) − m 2 1 − z 2 ] ⋅ Θ k m ( z ) = 0 , k ∈ ( 0.. ∞ ) . (1 - z^2) \cdot \frac{\displaystyle d^2 \Theta_{km}(z)}{\displaystyle dz^2} - 2 \cdot z \cdot \frac{\displaystyle d \Theta_{km}(z)}{\displaystyle dz} + \left[ k \cdot (k + 1) - \frac{\displaystyle m^2}{\displaystyle 1 - z^2} \right] \cdot \Theta_{km}(z) = 0, \quad k \in (0..\infty). ( 1 − z 2 ) ⋅ d z 2 d 2 Θ k m ( z ) − 2 ⋅ z ⋅ d z d Θ k m ( z ) + [ k ⋅ ( k + 1 ) − 1 − z 2 m 2 ] ⋅ Θ k m ( z ) = 0 , k ∈ ( 0..∞ ) . m = 0 m = 0 m = 0 P k ( z ) P_k(z) P k ( z )
которое продифференцируем m m m
d m d z m ( u ( z ) ⋅ v ( z ) ) = ∑ i = 0 m ( m i ) ⋅ d m − i d z m − i u ( z ) ⋅ d i d z i v ( z ) , \frac{\displaystyle d^m}{\displaystyle dz^m} \left( u(z) \cdot v(z) \right) = \sum_{i=0}^m \binom{m}{i} \cdot \frac{\displaystyle d^{m-i}}{\displaystyle dz^{m-i}} u(z) \cdot \frac{\displaystyle d^i}{\displaystyle dz^i} v(z), d z m d m ( u ( z ) ⋅ v ( z ) ) = i = 0 ∑ m ( i m ) ⋅ d z m − i d m − i u ( z ) ⋅ d z i d i v ( z ) , где ( m i ) = m ! i ! ⋅ ( m − i ) ! \binom{m}{i} = \frac{\displaystyle m!}{\displaystyle i! \cdot (m - i)!} ( i m ) = i ! ⋅ ( m − i )! m !
Начнём с первого члена уравнения (E.5 ( 1 − z 2 ) ⋅ d 2 P k ( z ) d z 2 − 2 ⋅ z ⋅ d P k ( z ) d z + k ⋅ ( k + 1 ) ⋅ P k ( z ) = 0 , (1 - z^2) \cdot \frac{\displaystyle d^2 P_k(z)}{\displaystyle dz^2} - 2 \cdot z \cdot \frac{\displaystyle d P_k(z)}{\displaystyle dz} + k \cdot (k + 1) \cdot P_k(z) = 0, ( 1 − z 2 ) ⋅ d z 2 d 2 P k ( z ) − 2 ⋅ z ⋅ d z d P k ( z ) + k ⋅ ( k + 1 ) ⋅ P k ( z ) = 0 , v ( z ) = 1 − z 2 v(z) = 1 - z^2 v ( z ) = 1 − z 2
d 0 d z 0 v = 1 − z 2 ; d 1 d z 1 v = − 2 ⋅ z ; d 2 d z 2 v = − 2 ; d i d z i v = 0 , i ∈ ( 3.. ∞ ) , \frac{\displaystyle d^0}{\displaystyle dz^0} v = 1 - z^2; \quad \frac{\displaystyle d^1}{\displaystyle dz^1} v = -2 \cdot z; \quad \frac{\displaystyle d^2}{\displaystyle dz^2} v = -2; \quad \frac{\displaystyle d^i}{\displaystyle dz^i} v = 0, \quad i \in (3..\infty), d z 0 d 0 v = 1 − z 2 ; d z 1 d 1 v = − 2 ⋅ z ; d z 2 d 2 v = − 2 ; d z i d i v = 0 , i ∈ ( 3..∞ ) , теперь примем v ( z ) = − 2 ⋅ z v(z) = -2 \cdot z v ( z ) = − 2 ⋅ z
d 0 d z 0 v = − 2 ⋅ z ; d 1 d z 1 v = − 2 ; d i d z i v = 0 , i ∈ ( 2.. ∞ ) , \frac{\displaystyle d^0}{\displaystyle dz^0} v = -2 \cdot z; \quad \frac{\displaystyle d^1}{\displaystyle dz^1} v = -2; \quad \frac{\displaystyle d^i}{\displaystyle dz^i} v = 0, \quad i \in (2..\infty), d z 0 d 0 v = − 2 ⋅ z ; d z 1 d 1 v = − 2 ; d z i d i v = 0 , i ∈ ( 2..∞ ) , для случая v ( z ) = k ⋅ ( k + 1 ) v(z) = k \cdot (k + 1) v ( z ) = k ⋅ ( k + 1 )
d 0 d z 0 v = k ⋅ ( k + 1 ) ; d i d z i v = 0 , i ∈ ( 1.. ∞ ) . \frac{\displaystyle d^0}{\displaystyle dz^0} v = k \cdot (k + 1); \quad \frac{\displaystyle d^i}{\displaystyle dz^i} v = 0, \quad i \in (1..\infty). d z 0 d 0 v = k ⋅ ( k + 1 ) ; d z i d i v = 0 , i ∈ ( 1..∞ ) . Биномиальные коэффициенты соответственно равны
( m 0 ) = 1 ; ( m 1 ) = m ; ( m 2 ) = m ⋅ ( m − 1 ) 2 . \binom{m}{0} = 1; \quad \binom{m}{1} = m; \quad \binom{m}{2} = \frac{\displaystyle m \cdot (m - 1)}{\displaystyle 2}. ( 0 m ) = 1 ; ( 1 m ) = m ; ( 2 m ) = 2 m ⋅ ( m − 1 ) . Результат дифференцирования первого члена уравнения (E.5 ( 1 − z 2 ) ⋅ d 2 P k ( z ) d z 2 − 2 ⋅ z ⋅ d P k ( z ) d z + k ⋅ ( k + 1 ) ⋅ P k ( z ) = 0 , (1 - z^2) \cdot \frac{\displaystyle d^2 P_k(z)}{\displaystyle dz^2} - 2 \cdot z \cdot \frac{\displaystyle d P_k(z)}{\displaystyle dz} + k \cdot (k + 1) \cdot P_k(z) = 0, ( 1 − z 2 ) ⋅ d z 2 d 2 P k ( z ) − 2 ⋅ z ⋅ d z d P k ( z ) + k ⋅ ( k + 1 ) ⋅ P k ( z ) = 0 ,
( 1 − z 2 ) ⋅ d m + 2 d z m + 2 P k ( z ) − 2 ⋅ z ⋅ m ⋅ d m + 1 d z m + 1 P k ( z ) − m ⋅ ( m − 1 ) ⋅ d m d z m P k ( z ) , (1 - z^2) \cdot \frac{\displaystyle d^{m+2}}{\displaystyle dz^{m+2}} P_k(z) - 2 \cdot z \cdot m \cdot \frac{\displaystyle d^{m+1}}{\displaystyle dz^{m+1}} P_k(z) - m \cdot (m - 1) \cdot \frac{\displaystyle d^m}{\displaystyle dz^m} P_k(z), ( 1 − z 2 ) ⋅ d z m + 2 d m + 2 P k ( z ) − 2 ⋅ z ⋅ m ⋅ d z m + 1 d m + 1 P k ( z ) − m ⋅ ( m − 1 ) ⋅ d z m d m P k ( z ) , для второго члена
− 2 ⋅ z ⋅ d m + 1 d z m + 1 P k ( z ) − 2 ⋅ m ⋅ d m d z m P k ( z ) , -2 \cdot z \cdot \frac{\displaystyle d^{m+1}}{\displaystyle dz^{m+1}} P_k(z) - 2 \cdot m \cdot \frac{\displaystyle d^m}{\displaystyle dz^m} P_k(z), − 2 ⋅ z ⋅ d z m + 1 d m + 1 P k ( z ) − 2 ⋅ m ⋅ d z m d m P k ( z ) , для третьего члена
k ⋅ ( k + 1 ) ⋅ d m d z m P k ( z ) , k \cdot (k + 1) \cdot \frac{\displaystyle d^m}{\displaystyle dz^m} P_k(z), k ⋅ ( k + 1 ) ⋅ d z m d m P k ( z ) , сложим все три члена
( 1 − z 2 ) ⋅ d m + 2 d z m + 2 P k ( z ) − 2 ⋅ ( m + 1 ) ⋅ z ⋅ d m + 1 d z m + 1 P k ( z ) + ( k − m ) ⋅ ( k + m + 1 ) ⋅ d m d z m P k ( z ) . (1 - z^2) \cdot \frac{\displaystyle d^{m+2}}{\displaystyle dz^{m+2}} P_k(z) - 2 \cdot (m + 1) \cdot z \cdot \frac{\displaystyle d^{m+1}}{\displaystyle dz^{m+1}} P_k(z) + (k-m) \cdot (k + m + 1) \cdot \frac{\displaystyle d^m}{\displaystyle dz^m} P_k(z). ( 1 − z 2 ) ⋅ d z m + 2 d m + 2 P k ( z ) − 2 ⋅ ( m + 1 ) ⋅ z ⋅ d z m + 1 d m + 1 P k ( z ) + ( k − m ) ⋅ ( k + m + 1 ) ⋅ d z m d m P k ( z ) . Левая часть уравнения (E.5 ( 1 − z 2 ) ⋅ d 2 P k ( z ) d z 2 − 2 ⋅ z ⋅ d P k ( z ) d z + k ⋅ ( k + 1 ) ⋅ P k ( z ) = 0 , (1 - z^2) \cdot \frac{\displaystyle d^2 P_k(z)}{\displaystyle dz^2} - 2 \cdot z \cdot \frac{\displaystyle d P_k(z)}{\displaystyle dz} + k \cdot (k + 1) \cdot P_k(z) = 0, ( 1 − z 2 ) ⋅ d z 2 d 2 P k ( z ) − 2 ⋅ z ⋅ d z d P k ( z ) + k ⋅ ( k + 1 ) ⋅ P k ( z ) = 0 , m m m
( 1 − z 2 ) ⋅ d m + 2 d z m + 2 P k ( z ) − 2 ⋅ ( m + 1 ) ⋅ z ⋅ d m + 1 d z m + 1 P k ( z ) + ( k − m ) ⋅ ( k + m + 1 ) ⋅ d m d z m P k ( z ) = 0. (1 - z^2) \cdot \frac{\displaystyle d^{m+2}}{\displaystyle dz^{m+2}} P_k(z) - 2 \cdot (m + 1) \cdot z \cdot \frac{\displaystyle d^{m+1}}{\displaystyle dz^{m+1}} P_k(z) + (k-m) \cdot (k + m + 1) \cdot \frac{\displaystyle d^m}{\displaystyle dz^m} P_k(z) = 0. ( 1 − z 2 ) ⋅ d z m + 2 d m + 2 P k ( z ) − 2 ⋅ ( m + 1 ) ⋅ z ⋅ d z m + 1 d m + 1 P k ( z ) + ( k − m ) ⋅ ( k + m + 1 ) ⋅ d z m d m P k ( z ) = 0. Полученное уравнение совпадает с уравнением (E.4 ( 1 − z 2 ) ⋅ d 2 Θ ^ k m ( z ) d z 2 − 2 ⋅ ( m + 1 ) ⋅ z ⋅ d Θ ^ k m ( z ) d z + ( k − m ) ⋅ ( k + m + 1 ) ⋅ Θ ^ k m ( z ) = 0. (1 - z^2) \cdot \frac{\displaystyle d^2 \widehat{\Theta}_{km}(z)}{\displaystyle dz^2} - 2 \cdot (m + 1) \cdot z \cdot \frac{\displaystyle d \widehat{\Theta}_{km}(z)}{\displaystyle dz} + (k - m) \cdot (k + m + 1) \cdot \widehat{\Theta}_{km}(z) = 0. ( 1 − z 2 ) ⋅ d z 2 d 2 Θ k m ( z ) − 2 ⋅ ( m + 1 ) ⋅ z ⋅ d z d Θ k m ( z ) + ( k − m ) ⋅ ( k + m + 1 ) ⋅ Θ k m ( z ) = 0. Θ ^ k m ( z ) = d m d z m P k ( z ) \widehat{\Theta}_{km}(z) = \frac{\displaystyle d^m}{\displaystyle dz^m} P_k(z) Θ k m ( z ) = d z m d m P k ( z ) E.1 d d z ( ( 1 − z 2 ) ⋅ d Θ ( z ) d z ) + [ γ 1 2 − m 2 1 − z 2 ] ⋅ Θ ( z ) = 0 , − 1 < z < 1 , \frac{\displaystyle d}{\displaystyle dz} \left( (1 - z^2) \cdot \frac{\displaystyle d \Theta(z)}{\displaystyle dz} \right) + \left[ \gamma_1^2 - \frac{\displaystyle m^2}{\displaystyle 1 - z^2} \right] \cdot \Theta(z) = 0, \quad -1 < z < 1, d z d ( ( 1 − z 2 ) ⋅ d z d Θ ( z ) ) + [ γ 1 2 − 1 − z 2 m 2 ] ⋅ Θ ( z ) = 0 , − 1 < z < 1 ,
где γ 1 k 2 = k ⋅ ( k + 1 ) \gamma_{1k}^2 = k \cdot (k + 1) γ 1 k 2 = k ⋅ ( k + 1 ) P k ( m ) ( z ) P_k^{(m)}(z) P k ( m ) ( z ) m > k m > k m > k d m d z m P k ( z ) \frac{\displaystyle d^m}{\displaystyle dz^m} P_k(z) d z m d m P k ( z ) k ≥ m k \ge m k ≥ m