Theorem: Hypergeometric Continued Fraction/Formulation 2

Let $a, b \in \C$.
Let $b \notin \Z_{\le 0}$.
:$\ds \frac {\map { {}_1 \operatorname F_1} {a + 1; b; x} } {\map { {}_1 \operatorname F_1} {a; b; x} } = \cfrac 1 {1 - \cfrac x {b + \cfrac {\paren {a + 1} x} {\paren {b + 1} + \cfrac {\paren {a - b } x} {\paren {b + 2} + \cfrac {\paren {a + 2} x} {\paren {b + 3} + \cfrac {\paren {a - b - 1 } x} {\paren {b + 4} + \cfrac {\ddots} {\ddots} } } } } } }$

Proof:
:First, we demonstate:
:$\ds \frac {\map { {}_1 \operatorname F_1} {a + 1; b; x} } {\map { {}_1 \operatorname F_1} {a; b; x} } = \frac 1 {1 - \dfrac x b \dfrac {\map { {}_1 \operatorname F_1} {a + 1; b + 1; x} } {\map { {}_1 \operatorname F_1} {a + 1; b; x} } }$
{{begin-eqn}}
{{eqn | l = \map { {}_1 \operatorname F_1} {a + 1; b; x} - \map { {}_1 \operatorname F_1} {a; b; x}
      | r = \frac x b \map { {}_1 \operatorname F_1} {a + 1; b + 1; x}
      | c = Hypergeometric Function Difference Equation/Formulation 4
}}
{{eqn | ll = \leadsto
      | l = \frac {\map { {}_1 \operatorname F_1} {a + 1; b; x} - \map { {}_1 \operatorname F_1} {a; b; x} } {\map { {}_1 \operatorname F_1} {a + 1; b; x} }
      | r = \frac x b \frac {\map { {}_1 \operatorname F_1} {a + 1; b + 1; x} } {\map { {}_1 \operatorname F_1} {a + 1; b; x} }
      | c = dividing both sides by $\map { {}_1 \operatorname F_1} {a + 1; b; x}$
}}
{{eqn | ll = \leadsto
      | l = \frac {\map { {}_1 \operatorname F_1} {a; b; x} } {\map { {}_1 \operatorname F_1} {a + 1; b; x} }
      | r = 1 - \frac x b \frac {\map { {}_1 \operatorname F_1} {a + 1; b + 1; x} } {\map { {}_1 \operatorname F_1} {a + 1; b; x} }
      | c = 
}}
{{eqn | ll = \leadsto
      | n = 1
      | l = \frac {\map { {}_1 \operatorname F_1} {a + 1; b; x} } {\map { {}_1 \operatorname F_1} {a; b; x} }
      | r = \frac 1 {1 - \dfrac x b \dfrac {\map { {}_1 \operatorname F_1} {a + 1; b + 1; x} } {\map { {}_1 \operatorname F_1} {a + 1; b; x} } }
      | c = taking the reciprocal of both sides
}}
{{end-eqn}}
:Next, we demonstate:
:$\ds \frac {\map { {}_1 \operatorname F_1} {a; b + 1; x} } {\map { {}_1 \operatorname F_1} {a; b; x} } = \frac 1 {1 + \dfrac {a x} {\paren {b + 1} b } \dfrac {\map { {}_1 \operatorname F_1} {a + 1; b + 2; x} } {\map { {}_1 \operatorname F_1} {a; b + 1; x} } }$
{{begin-eqn}}
{{eqn | l = \map { {}_1 \operatorname F_1} {a; b - 1; x} - \map { {}_1 \operatorname F_1} {a; b; x}
      | r = \frac {a x} {b \paren {b - 1} } \map { {}_1 \operatorname F_1} {a + 1; b + 1; x}
      | c = Hypergeometric Function Difference Equation/Formulation 3
}}
{{eqn | ll = \leadsto
      | l = \frac {\map { {}_1 \operatorname F_1} {a; b - 1; x} - \map { {}_1 \operatorname F_1} {a; b; x} } {\map { {}_1 \operatorname F_1} {a; b; x} }
      | r = \frac {a x} {b \paren {b - 1} } \frac {\map { {}_1 \operatorname F_1} {a + 1; b + 1; x} } {\map { {}_1 \operatorname F_1} {a; b; x} }
      | c = dividing both sides by $\map { {}_1 \operatorname F_1} {a; b; x}$
}}
{{eqn | ll = \leadsto
      | l = \frac {\map { {}_1 \operatorname F_1} {a; b - 1; x} } {\map { {}_1 \operatorname F_1} {a; b; x} }
      | r = 1 + \frac {a x} {b \paren {b - 1} } \frac {\map { {}_1 \operatorname F_1} {a + 1; b + 1; x} } {\map { {}_1 \operatorname F_1} {a; b; x} }
      | c = 
}}
{{eqn | ll = \leadsto
      | l = \frac {\map { {}_1 \operatorname F_1} {a; b; x} } {\map { {}_1 \operatorname F_1} {a; b - 1; x} }
      | r = \frac 1 {1 + \dfrac {a x} {b \paren {b - 1} } \dfrac {\map { {}_1 \operatorname F_1} {a + 1; b + 1; x} } {\map { {}_1 \operatorname F_1} {a; b; x} } }
      | c = taking the reciprocal of both sides
}}
{{eqn | ll = \leadsto
      | n = 2
      | l = \frac {\map { {}_1 \operatorname F_1} {a; b + 1; x} } {\map { {}_1 \operatorname F_1} {a; b; x} }
      | r = \frac 1 {1 + \dfrac {a x} {\paren {b + 1} b } \dfrac {\map { {}_1 \operatorname F_1} {a + 1; b + 2; x} } {\map { {}_1 \operatorname F_1} {a; b + 1; x} } }
      | c = $b \to b + 1$
}}
{{end-eqn}}
:Finally, we demonstate:
:$\ds \frac {\map { {}_1 \operatorname F_1} {a + 1; b + 1; x} } {\map { {}_1 \operatorname F_1} {a; b; x} } = \frac 1 {1 + \dfrac {\paren {a - b } x} {\paren {b + 1} b }  \dfrac {\map { {}_1 \operatorname F_1} {a + 1; b + 2; x} } {\map { {}_1 \operatorname F_1} {a + 1; b + 1; x} } }$
{{begin-eqn}}
{{eqn | l = \map { {}_1 \operatorname F_1} {a; b - 1; x} - \map { {}_1 \operatorname F_1} {a + 1; b; x}
      | r = \frac {\paren {a - b + 1} x} {b \paren {b - 1} } \map { {}_1 \operatorname F_1} {a + 1; b + 1; x}
      | c = Hypergeometric Function Difference Equation/Formulation 2
}}
{{eqn | ll = \leadsto
      | l = \frac {\map { {}_1 \operatorname F_1} {a; b - 1; x} - \map { {}_1 \operatorname F_1} {a + 1; b; x} } {\map { {}_1 \operatorname F_1} {a + 1; b; x} }
      | r = \frac {\paren {a - b + 1} x} {b \paren {b - 1} }  \frac {\map { {}_1 \operatorname F_1} {a + 1; b + 1; x} } {\map { {}_1 \operatorname F_1} {a + 1; b; x} }
      | c = dividing both sides by $\map { {}_1 \operatorname F_1} {a + 1; b; x}$
}}
{{eqn | ll = \leadsto
      | l = \frac {\map { {}_1 \operatorname F_1} {a; b - 1; x} } {\map { {}_1 \operatorname F_1} {a + 1; b; x} }
      | r = 1 + \frac {\paren {a - b + 1} x} {b \paren {b - 1} }  \frac {\map { {}_1 \operatorname F_1} {a + 1; b + 1; x} } {\map { {}_1 \operatorname F_1} {a + 1; b; x} }
      | c = 
}}
{{eqn | ll = \leadsto
      | l = \frac {\map { {}_1 \operatorname F_1} {a + 1; b; x} } {\map { {}_1 \operatorname F_1} {a; b - 1; x} }
      | r = \frac 1 {1 + \dfrac {\paren {a - b + 1} x} {b \paren {b - 1} }  \dfrac {\map { {}_1 \operatorname F_1} {a + 1; b + 1; x} } {\map { {}_1 \operatorname F_1} {a + 1; b; x} } }
      | c = taking the reciprocal of both sides
}}
{{eqn | ll = \leadsto
      | n = 3
      | l = \frac {\map { {}_1 \operatorname F_1} {a + 1; b + 1; x} } {\map { {}_1 \operatorname F_1} {a; b; x} }
      | r = \frac 1 {1 + \dfrac {\paren {a - b } x} {\paren {b + 1} b }  \dfrac {\map { {}_1 \operatorname F_1} {a + 1; b + 2; x} } {\map { {}_1 \operatorname F_1} {a + 1; b + 1; x} } }
      | c = $b \to b + 1$
}}
{{end-eqn}}
Therefore:
{{begin-eqn}}
{{eqn | l = \frac {\map { {}_1 \operatorname F_1} {a + 1; b; x} } {\map { {}_1 \operatorname F_1} {a; b; x} }
      | r = \frac 1 {1 - \dfrac x b \dfrac {\map { {}_1 \operatorname F_1} {a + 1; b + 1; x} } {\map { {}_1 \operatorname F_1} {a + 1; b; x} } }
      | c = from $\paren {1 }$ above
}}
{{eqn | ll = \leadsto
      | r = \cfrac 1 {1 - \dfrac x b \times \dfrac 1 {\paren {1 + \dfrac {\paren {a + 1} x} {\paren {b + 1} b } \dfrac {\map { {}_1 \operatorname F_1} {a + 2; b + 2; x} } {\map { {}_1 \operatorname F_1} {a + 1; b + 1; x} } } } }
      | c = from $\paren {2 }$ above
}}
{{eqn | ll = \leadsto
      | r = \cfrac 1 {1 - \dfrac x {b + \dfrac {\paren {a + 1} x} {\paren {b + 1} } \dfrac {\map { {}_1 \operatorname F_1} {a + 2; b + 2; x} } {\map { {}_1 \operatorname F_1} {a + 1; b + 1; x} } } }
      | c = 
}}
{{eqn | ll = \leadsto
      | r = \cfrac 1 {1 - \dfrac x {b + \dfrac {\paren {a + 1} x} {\paren {b + 1} } \times \dfrac 1 {\paren {1 + \dfrac {\paren {a - b } x} {\paren {b + 2} \paren {b + 1} }  \dfrac {\map { {}_1 \operatorname F_1} {a + 2; b + 3; x} } {\map { {}_1 \operatorname F_1} {a + 2; b + 2; x} } } } } }
      | c = from $\paren {3 }$ above
}}
{{eqn | ll = \leadsto
      | r = \cfrac 1 {1 - \dfrac x {b + \dfrac {\paren {a + 1} x} {\paren {b + 1} + \dfrac {\paren {a - b } x} {\paren {b + 2} }  \dfrac {\map { {}_1 \operatorname F_1} {a + 2; b + 3; x} } {\map { {}_1 \operatorname F_1} {a + 2; b + 2; x} } } } }
      | c = 
}}
{{eqn | ll = \leadsto
      | r = \cfrac 1 {1 - \dfrac x {b + \dfrac {\paren {a + 1} x} {\paren {b + 1} + \dfrac {\paren {a - b } x} {\paren {b + 2} }  \times \dfrac 1 {\paren {1 + \dfrac {\paren {a + 2} x} {\paren {b + 3} \paren {b + 2} } \dfrac {\map { {}_1 \operatorname F_1} {a + 3; b + 4; x} } {\map { {}_1 \operatorname F_1} {a + 2; b + 3; x} } } } } } }
      | c = from $\paren {2 }$ above
}}
{{eqn | ll = \leadsto
      | r = \cfrac 1 {1 - \dfrac x {b + \dfrac {\paren {a + 1} x} {\paren {b + 1} + \dfrac {\paren {a - b } x} {\paren {b + 2} + \dfrac {\paren {a + 2} x} {\paren {b + 3} } \dfrac {\map { {}_1 \operatorname F_1} {a + 3; b + 4; x} } {\map { {}_1 \operatorname F_1} {a + 2; b + 3; x} } } } } }
      | c = 
}}
{{eqn | ll = \leadsto
      | r = \cfrac 1 {1 - \dfrac x {b + \dfrac {\paren {a + 1} x} {\paren {b + 1} + \dfrac {\paren {a - b } x} {\paren {b + 2} + \dfrac {\paren {a + 2} x} {\paren {b + 3} } \times \dfrac 1 {\paren {1 + \dfrac {\paren {a - b - 1 } x} {\paren {b + 4} \paren {b + 3} }  \dfrac {\map { {}_1 \operatorname F_1} {a + 3; b + 5; x} } {\map { {}_1 \operatorname F_1} {a + 3; b + 4; x} } } } } } } }
      | c = from $\paren {3 }$ above
}}
{{eqn | ll = \leadsto
      | r = \cfrac 1 {1 - \cfrac x {b + \cfrac {\paren {a + 1} x} {\paren {b + 1} + \cfrac {\paren {a - b } x} {\paren {b + 2} + \cfrac {\paren {a + 2} x} {\paren {b + 3} + \cfrac {\paren {a - b - 1 } x} {\paren {b + 4} }  \cfrac {\map { {}_1 \operatorname F_1} {a + 3; b + 5; x} } {\map { {}_1 \operatorname F_1} {a + 3; b + 4; x} } } } } } }
      | c = 
}}
{{eqn | ll = \leadsto
      | r = \cdots
      | c = {{adinfinitum}}
}}
{{eqn | ll = \leadsto
      | r = \cfrac 1 {1 - \cfrac x {b + \cfrac {\paren {a + 1} x} {\paren {b + 1} + \cfrac {\paren {a - b } x} {\paren {b + 2} + \cfrac {\paren {a + 2} x} {\paren {b + 3} + \cfrac {\paren {a - b - 1 } x} {\paren {b + 4} + \cfrac {\ddots} {\ddots} } } } } } }
      | c = 
}}
{{end-eqn}}
{{qed}}
Category:Continued Fractions
Category:Hypergeometric Functions
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