In 1801 Leonhard Euler gave  the solution to the hypergeometric differential equation. It is closely related to the Gaussian hypergeometric function  , which was first systematically investigated by Carl Friedrich Gauß  .
Hypergeometric differential equation  
The hypergeometric function   , where the gamma function  denotes, satisfies the  linear differential equation of the 2nd order  :
  
    
      
        
          
            
             
            
              2 
             
           
          
            F. 
            
              1 
             
           
          ( 
          a 
          , 
          b 
          ; 
          c 
          ; 
          z 
          ) 
          = 
          
            ∑ 
            
              k 
              = 
              0 
             
            
              ∞ 
             
           
          
            
              
                Γ 
                ( 
                a 
                + 
                k 
                ) 
                 
                Γ 
                ( 
                b 
                + 
                k 
                ) 
                 
                Γ 
                ( 
                c 
                ) 
               
              
                Γ 
                ( 
                a 
                ) 
                 
                Γ 
                ( 
                b 
                ) 
                 
                Γ 
                ( 
                c 
                + 
                k 
                ) 
               
             
           
          
            
              
                z 
                
                  k 
                 
               
              
                k 
                ! 
               
             
           
         
       
     
    {\ displaystyle \ textstyle {} _ {2} F_ {1} (a, b; c; z) = \ sum _ {k = 0} ^ {\ infty} {\ frac {\ Gamma (a + k) \ , \ Gamma (b + k) \, \ Gamma (c)} {\ Gamma (a) \, \ Gamma (b) \, \ Gamma (c + k)}} {\ frac {z ^ {k}} {k!}}} 
   
 
  
    
      
        Γ 
        ( 
        ⋅ 
        ) 
       
     
    {\ displaystyle \ Gamma (\ cdot)} 
   
  
  
    
      
        z 
        ( 
        1 
        - 
        z 
        ) 
        
          
            
              
                d 
                
                  2 
                 
               
             
            
              
                
                  d 
                 
               
              
                z 
                
                  2 
                 
               
             
           
         
         
        
          
           
          
            2 
           
         
        
          F. 
          
            1 
           
         
        ( 
        a 
        , 
        b 
        ; 
        c 
        ; 
        z 
        ) 
        + 
        
          [ 
          
            c 
            - 
            ( 
            a 
            + 
            b 
            + 
            1 
            ) 
            z 
           
          ] 
         
        
          
            
              d 
             
            
              
                
                  d 
                 
               
              z 
             
           
         
         
        
          
           
          
            2 
           
         
        
          F. 
          
            1 
           
         
        ( 
        a 
        , 
        b 
        ; 
        c 
        ; 
        z 
        ) 
        - 
        a 
        b 
         
        
          
           
          
            2 
           
         
        
          F. 
          
            1 
           
         
        ( 
        a 
        , 
        b 
        ; 
        c 
        ; 
        z 
        ) 
        = 
        0 
       
     
    {\ displaystyle z (1-z) {\ frac {\ rm {d ^ {2}}} {{\ rm {d}} z ^ {2}}} \; {} _ {2} F_ {1} (a, b; c; z) + \ left [c- (a + b + 1) z \ right] {\ frac {\ rm {d}} {{\ rm {d}} z}} \; { } _ {2} F_ {1} (a, b; c; z) -ab \; {} _ {2} F_ {1} (a, b; c; z) = 0} 
   
  .  
Singularities 
 
The differential equation of the 2nd order has three liftable   singularities  , the values of which are determined in the following.
Based on the hypergeometric differential equation in the illustration
  
    
      
        
          
            
              
                d 
                
                  2 
                 
               
             
            
              
                
                  d 
                 
               
              
                z 
                
                  2 
                 
               
             
           
         
         
        
          
           
          
            2 
           
         
        
          F. 
          
            1 
           
         
        ( 
        a 
        , 
        b 
        ; 
        c 
        ; 
        z 
        ) 
        + 
        p 
        ( 
        z 
        ) 
        
          
            
              d 
             
            
              
                
                  d 
                 
               
              z 
             
           
         
         
        
          
           
          
            2 
           
         
        
          F. 
          
            1 
           
         
        ( 
        a 
        , 
        b 
        ; 
        c 
        ; 
        z 
        ) 
        - 
        q 
        ( 
        z 
        ) 
         
        
          
           
          
            2 
           
         
        
          F. 
          
            1 
           
         
        ( 
        a 
        , 
        b 
        ; 
        c 
        ; 
        z 
        ) 
        = 
        0 
       
     
    {\ displaystyle {\ frac {\ rm {d ^ {2}}} {{\ rm {d}} z ^ {2}}} \; {} _ {2} F_ {1} (a, b; c ; z) + p (z) {\ frac {\ rm {d}} {{\ rm {d}} z}} \; {} _ {2} F_ {1} (a, b; c; z) -q (z) \; {} _ {2} F_ {1} (a, b; c; z) = 0} 
   
  
With
  
    
      
        p 
        ( 
        z 
        ) 
        = 
        
          
            
              c 
              - 
              ( 
              a 
              + 
              b 
              + 
              1 
              ) 
              z 
             
            
              z 
              ( 
              1 
              - 
              z 
              ) 
             
           
         
        = 
        
          
            
              c 
              - 
              c 
              z 
              + 
              ( 
              c 
              - 
              a 
              - 
              b 
              - 
              1 
              ) 
              z 
             
            
              z 
              ( 
              1 
              - 
              z 
              ) 
             
           
         
        = 
        
          
            c 
            z 
           
         
        + 
        
          
            
              c 
              - 
              a 
              - 
              b 
              - 
              1 
             
            
              1 
              - 
              z 
             
           
         
       
     
    {\ displaystyle p (z) = {\ frac {c- (a + b + 1) z} {z (1-z)}} = {\ frac {c-cz + (cab-1) z} {z ( 1-z)}} = {\ frac {c} {z}} + {\ frac {cab-1} {1-z}}} 
   
  
and
  
    
      
        q 
        ( 
        z 
        ) 
        = 
        
          
            
              a 
              b 
             
            
              z 
              ( 
              1 
              - 
              z 
              ) 
             
           
         
       
     
    {\ displaystyle q (z) = {\ frac {ab} {z (1-z)}}} 
   
  
we get the two liftable   singularities  at and .
  
    
      
        z 
        = 
        0 
       
     
    {\ displaystyle z = 0} 
   
 
  
    
      
        z 
        = 
        1 
       
     
    {\ displaystyle z = 1} 
   
 
The third liftable singularity is obtained through the substitution   . First, the derivatives of the hypergeometric function are substituted as follows:
  
    
      
        
          t 
          = 
          
            
              1 
              z 
             
           
          , 
          
            
              
                
                  
                    d 
                   
                 
                t 
               
              
                
                  
                    d 
                   
                 
                z 
               
             
           
          = 
          
            
              
                - 
                1 
               
              
                z 
                
                  2 
                 
               
             
           
          = 
          - 
          
            t 
            
              2 
             
           
         
       
     
    {\ displaystyle \ textstyle t = {\ frac {1} {z}}, {\ frac {{\ rm {d}} t} {{\ rm {d}} z}} = {\ frac {-1} {z ^ {2}}} = - t ^ {2}} 
   
 
  
    
      
        
          
            
              d 
             
            
              
                
                  d 
                 
               
              z 
             
           
         
         
        
          
           
          
            2 
           
         
        
          F. 
          
            1 
           
         
        ( 
        a 
        , 
        b 
        ; 
        c 
        ; 
        z 
        ) 
        = 
        
          
            
              d 
             
            
              
                
                  d 
                 
               
              t 
             
           
         
         
        
          
           
          
            2 
           
         
        
          F. 
          
            1 
           
         
        ( 
        a 
        , 
        b 
        ; 
        c 
        ; 
        t 
        ) 
        ⋅ 
        
          
            
              
                
                  d 
                 
               
              t 
             
            
              
                
                  d 
                 
               
              z 
             
           
         
        = 
        - 
        
          t 
          
            2 
           
         
        ⋅ 
        
          
            
              d 
             
            
              
                
                  d 
                 
               
              t 
             
           
         
         
        
          
           
          
            2 
           
         
        
          F. 
          
            1 
           
         
        ( 
        a 
        , 
        b 
        ; 
        c 
        ; 
        t 
        ) 
       
     
    {\ displaystyle {\ frac {\ rm {d}} {{\ rm {d}} z}} \; {} _ {2} F_ {1} (a, b; c; z) = {\ frac { \ rm {d}} {{\ rm {d}} t}} \; {} _ {2} F_ {1} (a, b; c; t) \ cdot {\ frac {{\ rm {d} } t} {{\ rm {d}} z}} = - t ^ {2} \ cdot {\ frac {\ rm {d}} {{\ rm {d}} t}} \; {} _ { 2} F_ {1} (a, b; c; t)} 
   
  
and
  
    
      
        
          
            
              
                
                  
                    
                      
                        d 
                        
                          2 
                         
                       
                     
                    
                      
                        
                          d 
                         
                       
                      
                        z 
                        
                          2 
                         
                       
                     
                   
                 
                 
                
                  
                   
                  
                    2 
                   
                 
                
                  F. 
                  
                    1 
                   
                 
                ( 
                a 
                , 
                b 
                ; 
                c 
                ; 
                z 
                ) 
               
              
                 
                = 
                
                  
                    
                      d 
                     
                    
                      
                        
                          d 
                         
                       
                      t 
                     
                   
                 
                
                  
                    ( 
                   
                 
                - 
                
                  t 
                  
                    2 
                   
                 
                ⋅ 
                
                  
                    
                      d 
                     
                    
                      
                        
                          d 
                         
                       
                      t 
                     
                   
                 
                 
                
                  
                   
                  
                    2 
                   
                 
                
                  F. 
                  
                    1 
                   
                 
                ( 
                a 
                , 
                b 
                ; 
                c 
                ; 
                t 
                ) 
                
                  
                    ) 
                   
                 
                ⋅ 
                
                  
                    
                      
                        
                          d 
                         
                       
                      t 
                     
                    
                      
                        
                          d 
                         
                       
                      z 
                     
                   
                 
               
             
            
               
              
                 
                = 
                - 
                
                  t 
                  
                    2 
                   
                 
                
                  
                    ( 
                   
                 
                - 
                2 
                t 
                ⋅ 
                
                  
                    
                      d 
                     
                    
                      
                        
                          d 
                         
                       
                      t 
                     
                   
                 
                 
                
                  
                   
                  
                    2 
                   
                 
                
                  F. 
                  
                    1 
                   
                 
                ( 
                a 
                , 
                b 
                ; 
                c 
                ; 
                t 
                ) 
                - 
                
                  t 
                  
                    2 
                   
                 
                ⋅ 
                
                  
                    
                      
                        d 
                        
                          2 
                         
                       
                     
                    
                      
                        
                          d 
                         
                       
                      
                        t 
                        
                          2 
                         
                       
                     
                   
                 
                 
                
                  
                   
                  
                    2 
                   
                 
                
                  F. 
                  
                    1 
                   
                 
                ( 
                a 
                , 
                b 
                ; 
                c 
                ; 
                t 
                ) 
                
                  
                    ) 
                   
                 
               
             
            
               
              
                 
                = 
                
                  t 
                  
                    4th 
                   
                 
                ⋅ 
                
                  
                    
                      
                        d 
                        
                          2 
                         
                       
                     
                    
                      
                        
                          d 
                         
                       
                      
                        t 
                        
                          2 
                         
                       
                     
                   
                 
                 
                
                  
                   
                  
                    2 
                   
                 
                
                  F. 
                  
                    1 
                   
                 
                ( 
                a 
                , 
                b 
                ; 
                c 
                ; 
                t 
                ) 
                + 
                2 
                
                  t 
                  
                    3 
                   
                 
                ⋅ 
                
                  
                    
                      d 
                     
                    
                      
                        
                          d 
                         
                       
                      t 
                     
                   
                 
                 
                
                  
                   
                  
                    2 
                   
                 
                
                  F. 
                  
                    1 
                   
                 
                ( 
                a 
                , 
                b 
                ; 
                c 
                ; 
                t 
                ) 
               
             
           
         
       
     
    {\ displaystyle {\ begin {aligned} {\ frac {\ rm {d ^ {2}}} {{\ rm {d}} z ^ {2}}} \; {} _ {2} F_ {1} (a, b; c; z) & = {\ frac {\ rm {d}} {{\ rm {d}} t}} {\ Big (} -t ^ {2} \ cdot {\ frac {\ rm {d}} {{\ rm {d}} t}} \; {} _ {2} F_ {1} (a, b; c; t) {\ Big)} \ cdot {\ frac {{\ rm {d}} t} {{\ rm {d}} z}} \\ & = - t ^ {2} {\ Big (} -2t \ cdot {\ frac {\ rm {d}} {{\ rm {d}} t}} \; {} _ {2} F_ {1} (a, b; c; t) -t ^ {2} \ cdot {\ frac {\ rm {d ^ {2}} } {{\ rm {d}} t ^ {2}}} \; {} _ {2} F_ {1} (a, b; c; t) {\ Big)} \\ & = t ^ {4 } \ cdot {\ frac {\ rm {d ^ {2}}} {{\ rm {d}} t ^ {2}}} \; {} _ {2} F_ {1} (a, b; c ; t) + 2t ^ {3} \ cdot {\ frac {\ rm {d}} {{\ rm {d}} t}} \; {} _ {2} F_ {1} (a, b; c ; t) \ end {aligned}}} 
   
  
Thus the hypergeometric differential equation, after division by , takes on the following form:
  
    
      
        
          t 
          
            4th 
           
         
       
     
    {\ displaystyle t ^ {4}} 
   
 
  
    
      
        
          
            
              
                d 
                
                  2 
                 
               
             
            
              
                
                  d 
                 
               
              
                t 
                
                  2 
                 
               
             
           
         
         
        
          
           
          
            2 
           
         
        
          F. 
          
            1 
           
         
        ( 
        a 
        , 
        b 
        ; 
        c 
        ; 
        t 
        ) 
        + 
        
          
            
              p 
              ~ 
             
           
         
        ( 
        t 
        ) 
        ⋅ 
        
          
            
              d 
             
            
              
                
                  d 
                 
               
              t 
             
           
         
         
        
          
           
          
            2 
           
         
        
          F. 
          
            1 
           
         
        ( 
        a 
        , 
        b 
        ; 
        c 
        ; 
        t 
        ) 
        - 
        
          
            
              q 
              ~ 
             
           
         
        ( 
        t 
        ) 
         
        
          
           
          
            2 
           
         
        
          F. 
          
            1 
           
         
        ( 
        a 
        , 
        b 
        ; 
        c 
        ; 
        t 
        ) 
        = 
        0 
       
     
    {\ displaystyle {\ frac {\ rm {d ^ {2}}} {{\ rm {d}} t ^ {2}}} \; {} _ {2} F_ {1} (a, b; c ; t) + {\ tilde {p}} (t) \ cdot {\ frac {\ rm {d}} {{\ rm {d}} t}} \; {} _ {2} F_ {1} ( a, b; c; t) - {\ tilde {q}} (t) \; {} _ {2} F_ {1} (a, b; c; t) = 0} 
   
  
With
  
    
      
        
          
            
              p 
              ~ 
             
           
         
        ( 
        t 
        ) 
        = 
        
          
            2 
            t 
           
         
        + 
        
          
            1 
            
              t 
              
                2 
               
             
           
         
        p 
        ( 
        z 
        = 
        
          
            
              1 
              t 
             
           
         
        ) 
        = 
        
          
            2 
            t 
           
         
        + 
        
          
            1 
            
              t 
              
                2 
               
             
           
         
        
          
            ( 
           
         
        c 
        t 
        + 
        
          
            
              c 
              - 
              a 
              - 
              b 
              - 
              1 
             
            
              1 
              - 
              
                
                  1 
                  t 
                 
               
             
           
         
        
          
            ) 
           
         
        = 
        
          
            
              c 
              + 
              2 
             
            t 
           
         
        + 
        
          
            
              c 
              - 
              a 
              - 
              b 
              - 
              1 
             
            
              t 
              ( 
              t 
              - 
              1 
              ) 
             
           
         
       
     
    {\ displaystyle {\ tilde {p}} (t) = {\ frac {2} {t}} + {\ frac {1} {t ^ {2}}} p (z = {\ tfrac {1} { t}}) = {\ frac {2} {t}} + {\ frac {1} {t ^ {2}}} {\ Big (} ct + {\ frac {cab-1} {1 - {\ frac {1} {t}}}} {\ Big)} = {\ frac {c + 2} {t}} + {\ frac {cab-1} {t (t-1)}}} 
   
  
and
  
    
      
        
          
            
              q 
              ~ 
             
           
         
        ( 
        t 
        ) 
        = 
        
          
            1 
            
              t 
              
                4th 
               
             
           
         
        q 
        ( 
        z 
        = 
        
          
            
              1 
              t 
             
           
         
        ) 
        = 
        
          
            1 
            
              t 
              
                4th 
               
             
           
         
        
          
            
              a 
              b 
             
            
              
                
                  1 
                  t 
                 
               
              ( 
              1 
              - 
              
                
                  1 
                  t 
                 
               
              ) 
             
           
         
        = 
        
          
            
              a 
              b 
             
            
              
                t 
                
                  2 
                 
               
              ( 
              t 
              - 
              1 
              ) 
             
           
         
       
     
    {\ displaystyle {\ tilde {q}} (t) = {\ frac {1} {t ^ {4}}} q (z = {\ tfrac {1} {t}}) = {\ frac {1} {t ^ {4}}} {\ frac {ab} {{\ frac {1} {t}} (1 - {\ frac {1} {t}})}} = {\ frac {ab} {t ^ {2} (t-1)}}} 
   
  
Accordingly, the hypergeometric differential equation also has a lifting singularity at.
  
    
      
        z 
        = 
        
          
            
              1 
              t 
             
           
         
        = 
        ∞ 
       
     
    {\ displaystyle z = {\ tfrac {1} {t}} = \ infty} 
   
 
Solution of the hypergeometric differential equation 
 
With the power series approach   with complex coefficients , the hypergeometric differential equation reads:
  
    
      
        
          u 
          ( 
          z 
          ) 
          = 
          
            ∑ 
            
              k 
              = 
              0 
             
            
              ∞ 
             
           
          
            u 
            
              k 
             
           
          
            z 
            
              k 
             
           
         
       
     
    {\ displaystyle \ textstyle u (z) = \ sum _ {k = 0} ^ {\ infty} u_ {k} z ^ {k}} 
   
 
  
    
      
        
          u 
          
            k 
           
         
       
     
    {\ displaystyle u_ {k}} 
   
 
  
    
      
        z 
        ( 
        1 
        - 
        z 
        ) 
        
          
            
              
                d 
                
                  2 
                 
               
             
            
              
                
                  d 
                 
               
              
                z 
                
                  2 
                 
               
             
           
         
        
          ∑ 
          
            k 
            = 
            0 
           
          
            ∞ 
           
         
        
          u 
          
            k 
           
         
        
          z 
          
            k 
           
         
        + 
        
          [ 
          
            c 
            - 
            ( 
            a 
            + 
            b 
            + 
            1 
            ) 
            z 
           
          ] 
         
        
          
            
              d 
             
            
              
                
                  d 
                 
               
              z 
             
           
         
        
          ∑ 
          
            k 
            = 
            0 
           
          
            ∞ 
           
         
        
          u 
          
            k 
           
         
        
          z 
          
            k 
           
         
        - 
        a 
        b 
        
          ∑ 
          
            k 
            = 
            0 
           
          
            ∞ 
           
         
        
          u 
          
            k 
           
         
        
          z 
          
            k 
           
         
        = 
        0 
       
     
    {\ displaystyle z (1-z) {\ frac {\ rm {d ^ {2}}} {{\ rm {d}} z ^ {2}}} \ sum _ {k = 0} ^ {\ infty } u_ {k} z ^ {k} + \ left [c- (a + b + 1) z \ right] {\ frac {\ rm {d}} {{\ rm {d}} z}} \ sum _ {k = 0} ^ {\ infty} u_ {k} z ^ {k} -ab \ sum _ {k = 0} ^ {\ infty} u_ {k} z ^ {k} = 0} 
   
  .  
After executing the derivations, the representation results
  
    
      
        z 
        ( 
        1 
        - 
        z 
        ) 
        
          ∑ 
          
            k 
            = 
            2 
           
          
            ∞ 
           
         
        k 
        ( 
        k 
        - 
        1 
        ) 
        
          u 
          
            k 
           
         
        
          z 
          
            k 
            - 
            2 
           
         
        + 
        
          [ 
          
            c 
            - 
            ( 
            a 
            + 
            b 
            + 
            1 
            ) 
            z 
           
          ] 
         
        
          ∑ 
          
            k 
            = 
            1 
           
          
            ∞ 
           
         
        k 
        
          u 
          
            k 
           
         
        
          z 
          
            k 
            - 
            1 
           
         
        - 
        a 
        b 
        
          ∑ 
          
            k 
            = 
            0 
           
          
            ∞ 
           
         
        
          u 
          
            k 
           
         
        
          z 
          
            k 
           
         
        = 
        0 
       
     
    {\ displaystyle z (1-z) \ sum _ {k = 2} ^ {\ infty} k (k-1) u_ {k} z ^ {k-2} + \ left [c- (a + b + 1) z \ right] \ sum _ {k = 1} ^ {\ infty} ku_ {k} z ^ {k-1} -ab \ sum _ {k = 0} ^ {\ infty} u_ {k} z ^ {k} = 0} 
   
  .  
Summarizing the powers of leads to
  
    
      
        z 
       
     
    {\ displaystyle z} 
   
 
  
    
      
        
          ∑ 
          
            k 
            = 
            2 
           
          
            ∞ 
           
         
        k 
        ( 
        k 
        - 
        1 
        ) 
        
          u 
          
            k 
           
         
        
          z 
          
            k 
            - 
            1 
           
         
        - 
        
          ∑ 
          
            k 
            = 
            2 
           
          
            ∞ 
           
         
        k 
        ( 
        k 
        - 
        1 
        ) 
        
          u 
          
            k 
           
         
        
          z 
          
            k 
           
         
        + 
        c 
        
          ∑ 
          
            k 
            = 
            1 
           
          
            ∞ 
           
         
        k 
        
          u 
          
            k 
           
         
        
          z 
          
            k 
            - 
            1 
           
         
        - 
        ( 
        a 
        + 
        b 
        + 
        1 
        ) 
        
          ∑ 
          
            k 
            = 
            1 
           
          
            ∞ 
           
         
        k 
        
          u 
          
            k 
           
         
        
          z 
          
            k 
           
         
        - 
        a 
        b 
        
          ∑ 
          
            k 
            = 
            0 
           
          
            ∞ 
           
         
        
          u 
          
            k 
           
         
        
          z 
          
            k 
           
         
        = 
        0 
       
     
    {\ displaystyle \ sum _ {k = 2} ^ {\ infty} k (k-1) u_ {k} z ^ {k-1} - \ sum _ {k = 2} ^ {\ infty} k (k -1) u_ {k} z ^ {k} + c \ sum _ {k = 1} ^ {\ infty} ku_ {k} z ^ {k-1} - (a + b + 1) \ sum _ { k = 1} ^ {\ infty} ku_ {k} z ^ {k} -ab \ sum _ {k = 0} ^ {\ infty} u_ {k} z ^ {k} = 0} 
   
  .  
The index shift  results in
  
    
      
        
          ∑ 
          
            k 
            = 
            0 
           
          
            ∞ 
           
         
        ( 
        k 
        + 
        1 
        ) 
        k 
        
          u 
          
            k 
            + 
            1 
           
         
        
          z 
          
            k 
           
         
        - 
        
          ∑ 
          
            k 
            = 
            0 
           
          
            ∞ 
           
         
        k 
        ( 
        k 
        - 
        1 
        ) 
        
          u 
          
            k 
           
         
        
          z 
          
            k 
           
         
        + 
        c 
        
          ∑ 
          
            k 
            = 
            0 
           
          
            ∞ 
           
         
        ( 
        k 
        + 
        1 
        ) 
        
          u 
          
            k 
            + 
            1 
           
         
        
          z 
          
            k 
           
         
        - 
        ( 
        a 
        + 
        b 
        + 
        1 
        ) 
        
          ∑ 
          
            k 
            = 
            0 
           
          
            ∞ 
           
         
        k 
        
          u 
          
            k 
           
         
        
          z 
          
            k 
           
         
        - 
        a 
        b 
        
          ∑ 
          
            k 
            = 
            0 
           
          
            ∞ 
           
         
        
          u 
          
            k 
           
         
        
          z 
          
            k 
           
         
        = 
        0 
       
     
    {\ displaystyle \ sum _ {k = 0} ^ {\ infty} (k + 1) ku_ {k + 1} z ^ {k} - \ sum _ {k = 0} ^ {\ infty} k (k- 1) u_ {k} z ^ {k} + c \ sum _ {k = 0} ^ {\ infty} (k + 1) u_ {k + 1} z ^ {k} - (a + b + 1) \ sum _ {k = 0} ^ {\ infty} ku_ {k} z ^ {k} -ab \ sum _ {k = 0} ^ {\ infty} u_ {k} z ^ {k} = 0} 
   
  .  
This equation is obviously true if:
  
    
      
        ( 
        k 
        + 
        1 
        ) 
        k 
        
          u 
          
            k 
            + 
            1 
           
         
        - 
        k 
        ( 
        k 
        - 
        1 
        ) 
        
          u 
          
            k 
           
         
        + 
        c 
        ( 
        k 
        + 
        1 
        ) 
        
          u 
          
            k 
            + 
            1 
           
         
        - 
        ( 
        a 
        + 
        b 
        + 
        1 
        ) 
        k 
        
          u 
          
            k 
           
         
        - 
        a 
        b 
        
          u 
          
            k 
           
         
        = 
        0 
       
     
    {\ displaystyle (k + 1) ku_ {k + 1} -k (k-1) u_ {k} + c (k + 1) u_ {k + 1} - (a + b + 1) ku_ {k} -abu_ {k} = 0} 
   
  .  
The following recursion has thus been found for the coefficient :
  
    
      
        
          u 
          
            k 
           
         
       
     
    {\ displaystyle u_ {k}} 
   
 
  
    
      
        
          
            
              
                
                  u 
                  
                    k 
                    + 
                    1 
                   
                 
               
              
                 
                = 
                
                  
                    
                      k 
                      ( 
                      k 
                      - 
                      1 
                      ) 
                      + 
                      ( 
                      a 
                      + 
                      b 
                      + 
                      1 
                      ) 
                      k 
                      + 
                      a 
                      b 
                     
                    
                      ( 
                      k 
                      + 
                      1 
                      ) 
                      k 
                      + 
                      c 
                      ( 
                      k 
                      + 
                      1 
                      ) 
                     
                   
                 
                
                  u 
                  
                    k 
                   
                 
                 
                = 
                
                  
                    
                      
                        k 
                        
                          2 
                         
                       
                      - 
                      k 
                      + 
                      k 
                      a 
                      + 
                      k 
                      b 
                      + 
                      k 
                      + 
                      a 
                      b 
                     
                    
                      ( 
                      c 
                      + 
                      k 
                      ) 
                      ( 
                      1 
                      + 
                      k 
                      ) 
                     
                   
                 
                
                  u 
                  
                    k 
                   
                 
               
             
            
               
              
                 
                = 
                
                  
                    
                      
                        k 
                        
                          2 
                         
                       
                      + 
                      k 
                      a 
                      + 
                      k 
                      b 
                      + 
                      a 
                      b 
                     
                    
                      ( 
                      c 
                      + 
                      k 
                      ) 
                      ( 
                      1 
                      + 
                      k 
                      ) 
                     
                   
                 
                
                  u 
                  
                    k 
                   
                 
                 
                 
                 
                 
                 
                = 
                
                  
                    
                      ( 
                      a 
                      + 
                      k 
                      ) 
                      ( 
                      b 
                      + 
                      k 
                      ) 
                     
                    
                      ( 
                      c 
                      + 
                      k 
                      ) 
                      ( 
                      1 
                      + 
                      k 
                      ) 
                     
                   
                 
                
                  u 
                  
                    k 
                   
                 
               
              
                = 
                
                  
                    
                      ( 
                      a 
                      , 
                      k 
                      ) 
                      ( 
                      b 
                      , 
                      k 
                      ) 
                     
                    
                      ( 
                      c 
                      , 
                      k 
                      ) 
                      ( 
                      1 
                      , 
                      k 
                      ) 
                     
                   
                 
                
                  u 
                  
                    k 
                   
                 
               
             
           
         
       
     
    {\ displaystyle {\ begin {aligned} u_ {k + 1} & = {\ frac {k (k-1) + (a + b + 1) k + ab} {(k + 1) k + c (k +1)}} u_ {k} \ qquad = {\ frac {k ^ {2} -k + ka + kb + k + ab} {(c + k) (1 + k)}} u_ {k} \ \ & = {\ frac {k ^ {2} + ka + kb + ab} {(c + k) (1 + k)}} u_ {k} \ qquad \ qquad \ qquad \; \; = {\ frac {(a + k) (b + k)} {(c + k) (1 + k)}} u_ {k} & = {\ frac {(a, k) (b, k)} {(c, k) (1, k)}} u_ {k} \ end {aligned}}} 
   
  
Here the Pochhammer symbol  denotes .
  
    
      
        ( 
        x 
        , 
        n 
        ) 
        ≡ 
        
          
            
              
                Γ 
                ( 
                x 
                + 
                n 
                ) 
               
              
                Γ 
                ( 
                x 
                ) 
               
             
           
         
       
     
    {\ displaystyle (x, n) \ equiv {\ tfrac {\ Gamma (x + n)} {\ Gamma (x)}}} 
   
  
If the initial value is set, the first basic solution of the hypergeometric differential equation is:
  
    
      
        
          u 
          
            0 
           
         
        = 
        1 
       
     
    {\ displaystyle u_ {0} = 1} 
   
 
  
    
      
        u 
        ( 
        z 
        ) 
        = 
        
          
           
          
            2 
           
         
        
          F. 
          
            1 
           
         
        ( 
        a 
        , 
        b 
        ; 
        c 
        ; 
        z 
        ) 
        = 
        
          ∑ 
          
            k 
            = 
            0 
           
          
            ∞ 
           
         
        
          
            
              ( 
              a 
              , 
              k 
              ) 
              ( 
              b 
              , 
              k 
              ) 
             
            
              ( 
              c 
              , 
              k 
              ) 
              ( 
              1 
              , 
              k 
              ) 
             
           
         
        
          z 
          
            k 
           
         
        = 
        
          ∑ 
          
            k 
            = 
            0 
           
          
            ∞ 
           
         
        
          
            
              Γ 
              ( 
              a 
              + 
              k 
              ) 
               
              Γ 
              ( 
              b 
              + 
              k 
              ) 
               
              Γ 
              ( 
              c 
              ) 
             
            
              Γ 
              ( 
              a 
              ) 
               
              Γ 
              ( 
              b 
              ) 
               
              Γ 
              ( 
              c 
              + 
              k 
              ) 
             
           
         
        
          
            
              z 
              
                k 
               
             
            
              k 
              ! 
             
           
         
       
     
    {\ displaystyle u (z) = {} _ {2} F_ {1} (a, b; c; z) = \ sum _ {k = 0} ^ {\ infty} {\ frac {(a, k) (b, k)} {(c, k) (1, k)}} z ^ {k} = \ sum _ {k = 0} ^ {\ infty} {\ frac {\ Gamma (a + k) \ , \ Gamma (b + k) \, \ Gamma (c)} {\ Gamma (a) \, \ Gamma (b) \, \ Gamma (c + k)}} {\ frac {z ^ {k}} {k!}}} 
   
  .  
For is obtained as a second linearly independent basic solution
  
    
      
        c 
        ∉ 
        
          Z 
         
       
     
    {\ displaystyle c \ notin \ mathbb {Z}} 
   
 
  
    
      
        v 
        ( 
        z 
        ) 
        = 
        
          z 
          
            1 
            - 
            c 
           
         
        
          
           
          
            2 
           
         
        
          F. 
          
            1 
           
         
        ( 
        a 
        - 
        c 
        + 
        1 
        , 
        b 
        - 
        c 
        + 
        1 
        ; 
        2 
        - 
        c 
        ; 
        z 
        ) 
       
     
    {\ displaystyle v (z) = z ^ {1-c} {} _ {2} F_ {1} (a-c + 1, b-c + 1; 2-c; z)} 
   
  
Both together span the entire solution space of the hypergeometric differential equation:
  
    
      
        y 
        ( 
        z 
        ) 
        = 
        
          C. 
          
            1 
           
         
        u 
        ( 
        z 
        ) 
        + 
        
          C. 
          
            2 
           
         
        v 
        ( 
        z 
        ) 
       
     
    {\ displaystyle y (z) = C_ {1} u (z) + C_ {2} v (z)} 
   
   With 
  
    
      
        
          C. 
          
            1 
           
         
        , 
        
          C. 
          
            2 
           
         
        ∈ 
        
          C. 
         
       
     
    {\ displaystyle C_ {1}, C_ {2} \ in \ mathbb {C}} 
   
  
  
See also  
literature  
Individual evidence  
^    Leonhard Euler: Transformationis Singularis, Nova Acta Academiae Scientiarum Imperialis Petropolitanae, Volume 12, 1801, Pages 58-70, online   at books.google.de 
 
↑    Erwin Kreyszig: Advanced Engineering Mathematics, John Wiley & Sons 1988, page 204f. 
 
 
 
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