Lambert published several examples of continued fractions in this form in 1768, and both Euler and Lagrange investigated similar constructions,[ 1] Carl Friedrich Gauss who utilized the algebra described in the next section to deduce the general form of this continued fraction, in 1813.[ 2]
Although Gauss gave the form of this continued fraction, he did not give a proof of its convergence properties. Bernhard Riemann [ 3] [ 4] Edward Burr Van Vleck .[ 5]
Let
f
0
,
f
1
,
f
2
,
…
{\displaystyle f_{0},f_{1},f_{2},\dots }
three-term recurrence relation
f
i
−
1
=
f
i
+
k
i
z
f
i
+
1
{\displaystyle f_{i-1}=f_{i}+k_{i}\,z\,f_{i+1}}
for all
i
>
0
{\displaystyle i>0}
k
i
{\displaystyle k_{i}}
Then
f
i
−
1
f
i
=
1
+
k
i
z
f
i
+
1
f
i
,
and so
f
i
f
i
−
1
=
1
1
+
k
i
z
f
i
+
1
f
i
{\displaystyle {\frac {f_{i-1}}{f_{i}}}=1+k_{i}z{\frac {f_{i+1}}{f_{i}}},{\text{ and so }}{\frac {f_{i}}{f_{i-1}}}={\frac {1}{1+k_{i}z{\frac {f_{i+1}}{f_{i}}}}}}
Setting
g
i
=
f
i
/
f
i
−
1
,
{\displaystyle g_{i}=f_{i}/f_{i-1},}
g
i
=
1
1
+
k
i
z
g
i
+
1
,
{\displaystyle g_{i}={\frac {1}{1+k_{i}zg_{i+1}}},}
So
g
1
=
f
1
f
0
=
1
1
+
k
1
z
g
2
=
1
1
+
k
1
z
1
+
k
2
z
g
3
=
1
1
+
k
1
z
1
+
k
2
z
1
+
k
3
z
g
4
=
⋯
.
{\displaystyle g_{1}={\frac {f_{1}}{f_{0}}}={\cfrac {1}{1+k_{1}zg_{2}}}={\cfrac {1}{1+{\cfrac {k_{1}z}{1+k_{2}zg_{3}}}}}={\cfrac {1}{1+{\cfrac {k_{1}z}{1+{\cfrac {k_{2}z}{1+k_{3}zg_{4}}}}}}}=\cdots .\ }
Repeating this ad infinitum produces the continued fraction expression
f
1
f
0
=
1
1
+
k
1
z
1
+
k
2
z
1
+
k
3
z
1
+
⋱
{\displaystyle {\frac {f_{1}}{f_{0}}}={\cfrac {1}{1+{\cfrac {k_{1}z}{1+{\cfrac {k_{2}z}{1+{\cfrac {k_{3}z}{1+{}\ddots }}}}}}}}}
In Gauss's continued fraction, the functions
f
i
{\displaystyle f_{i}}
0
F
1
{\displaystyle {}_{0}F_{1}}
1
F
1
{\displaystyle {}_{1}F_{1}}
2
F
1
{\displaystyle {}_{2}F_{1}}
f
i
−
1
−
f
i
=
k
i
z
f
i
+
1
{\displaystyle f_{i-1}-f_{i}=k_{i}zf_{i+1}}
The simplest case involves
0
F
1
(
a
;
z
)
=
1
+
1
a
1
!
z
+
1
a
(
a
+
1
)
2
!
z
2
+
1
a
(
a
+
1
)
(
a
+
2
)
3
!
z
3
+
⋯
.
{\displaystyle \,_{0}F_{1}(a;z)=1+{\frac {1}{a\,1!}}z+{\frac {1}{a(a+1)\,2!}}z^{2}+{\frac {1}{a(a+1)(a+2)\,3!}}z^{3}+\cdots .}
Starting with the identity
0
F
1
(
a
−
1
;
z
)
−
0
F
1
(
a
;
z
)
=
z
a
(
a
−
1
)
0
F
1
(
a
+
1
;
z
)
,
{\displaystyle \,_{0}F_{1}(a-1;z)-\,_{0}F_{1}(a;z)={\frac {z}{a(a-1)}}\,_{0}F_{1}(a+1;z),}
we may take
f
i
=
0
F
1
(
a
+
i
;
z
)
,
k
i
=
1
(
a
+
i
)
(
a
+
i
−
1
)
,
{\displaystyle f_{i}={}_{0}F_{1}(a+i;z),\,k_{i}={\tfrac {1}{(a+i)(a+i-1)}},}
giving
0
F
1
(
a
+
1
;
z
)
0
F
1
(
a
;
z
)
=
1
1
+
1
a
(
a
+
1
)
z
1
+
1
(
a
+
1
)
(
a
+
2
)
z
1
+
1
(
a
+
2
)
(
a
+
3
)
z
1
+
⋱
{\displaystyle {\frac {\,_{0}F_{1}(a+1;z)}{\,_{0}F_{1}(a;z)}}={\cfrac {1}{1+{\cfrac {{\frac {1}{a(a+1)}}z}{1+{\cfrac {{\frac {1}{(a+1)(a+2)}}z}{1+{\cfrac {{\frac {1}{(a+2)(a+3)}}z}{1+{}\ddots }}}}}}}}}
or
0
F
1
(
a
+
1
;
z
)
a
0
F
1
(
a
;
z
)
=
1
a
+
z
(
a
+
1
)
+
z
(
a
+
2
)
+
z
(
a
+
3
)
+
⋱
.
{\displaystyle {\frac {\,_{0}F_{1}(a+1;z)}{a\,_{0}F_{1}(a;z)}}={\cfrac {1}{a+{\cfrac {z}{(a+1)+{\cfrac {z}{(a+2)+{\cfrac {z}{(a+3)+{}\ddots }}}}}}}}.}
This expansion converges to the meromorphic function defined by the ratio of the two convergent series (provided, of course, that a is neither zero nor a negative integer).
The next case involves
1
F
1
(
a
;
b
;
z
)
=
1
+
a
b
1
!
z
+
a
(
a
+
1
)
b
(
b
+
1
)
2
!
z
2
+
a
(
a
+
1
)
(
a
+
2
)
b
(
b
+
1
)
(
b
+
2
)
3
!
z
3
+
⋯
{\displaystyle {}_{1}F_{1}(a;b;z)=1+{\frac {a}{b\,1!}}z+{\frac {a(a+1)}{b(b+1)\,2!}}z^{2}+{\frac {a(a+1)(a+2)}{b(b+1)(b+2)\,3!}}z^{3}+\cdots }
for which the two identities
1
F
1
(
a
;
b
−
1
;
z
)
−
1
F
1
(
a
+
1
;
b
;
z
)
=
(
a
−
b
+
1
)
z
b
(
b
−
1
)
1
F
1
(
a
+
1
;
b
+
1
;
z
)
{\displaystyle \,_{1}F_{1}(a;b-1;z)-\,_{1}F_{1}(a+1;b;z)={\frac {(a-b+1)z}{b(b-1)}}\,_{1}F_{1}(a+1;b+1;z)}
1
F
1
(
a
;
b
−
1
;
z
)
−
1
F
1
(
a
;
b
;
z
)
=
a
z
b
(
b
−
1
)
1
F
1
(
a
+
1
;
b
+
1
;
z
)
{\displaystyle \,_{1}F_{1}(a;b-1;z)-\,_{1}F_{1}(a;b;z)={\frac {az}{b(b-1)}}\,_{1}F_{1}(a+1;b+1;z)}
are used alternately.
Let
f
0
(
z
)
=
1
F
1
(
a
;
b
;
z
)
,
{\displaystyle f_{0}(z)=\,_{1}F_{1}(a;b;z),}
f
1
(
z
)
=
1
F
1
(
a
+
1
;
b
+
1
;
z
)
,
{\displaystyle f_{1}(z)=\,_{1}F_{1}(a+1;b+1;z),}
f
2
(
z
)
=
1
F
1
(
a
+
1
;
b
+
2
;
z
)
,
{\displaystyle f_{2}(z)=\,_{1}F_{1}(a+1;b+2;z),}
f
3
(
z
)
=
1
F
1
(
a
+
2
;
b
+
3
;
z
)
,
{\displaystyle f_{3}(z)=\,_{1}F_{1}(a+2;b+3;z),}
f
4
(
z
)
=
1
F
1
(
a
+
2
;
b
+
4
;
z
)
,
{\displaystyle f_{4}(z)=\,_{1}F_{1}(a+2;b+4;z),}
etc.
This gives
f
i
−
1
−
f
i
=
k
i
z
f
i
+
1
{\displaystyle f_{i-1}-f_{i}=k_{i}zf_{i+1}}
k
1
=
a
−
b
b
(
b
+
1
)
,
k
2
=
a
+
1
(
b
+
1
)
(
b
+
2
)
,
k
3
=
a
−
b
−
1
(
b
+
2
)
(
b
+
3
)
,
k
4
=
a
+
2
(
b
+
3
)
(
b
+
4
)
{\displaystyle k_{1}={\tfrac {a-b}{b(b+1)}},k_{2}={\tfrac {a+1}{(b+1)(b+2)}},k_{3}={\tfrac {a-b-1}{(b+2)(b+3)}},k_{4}={\tfrac {a+2}{(b+3)(b+4)}}}
1
F
1
(
a
+
1
;
b
+
1
;
z
)
1
F
1
(
a
;
b
;
z
)
=
1
1
+
a
−
b
b
(
b
+
1
)
z
1
+
a
+
1
(
b
+
1
)
(
b
+
2
)
z
1
+
a
−
b
−
1
(
b
+
2
)
(
b
+
3
)
z
1
+
a
+
2
(
b
+
3
)
(
b
+
4
)
z
1
+
⋱
{\displaystyle {\frac {{}_{1}F_{1}(a+1;b+1;z)}{{}_{1}F_{1}(a;b;z)}}={\cfrac {1}{1+{\cfrac {{\frac {a-b}{b(b+1)}}z}{1+{\cfrac {{\frac {a+1}{(b+1)(b+2)}}z}{1+{\cfrac {{\frac {a-b-1}{(b+2)(b+3)}}z}{1+{\cfrac {{\frac {a+2}{(b+3)(b+4)}}z}{1+{}\ddots }}}}}}}}}}}
or
1
F
1
(
a
+
1
;
b
+
1
;
z
)
b
1
F
1
(
a
;
b
;
z
)
=
1
b
+
(
a
−
b
)
z
(
b
+
1
)
+
(
a
+
1
)
z
(
b
+
2
)
+
(
a
−
b
−
1
)
z
(
b
+
3
)
+
(
a
+
2
)
z
(
b
+
4
)
+
⋱
{\displaystyle {\frac {{}_{1}F_{1}(a+1;b+1;z)}{b{}_{1}F_{1}(a;b;z)}}={\cfrac {1}{b+{\cfrac {(a-b)z}{(b+1)+{\cfrac {(a+1)z}{(b+2)+{\cfrac {(a-b-1)z}{(b+3)+{\cfrac {(a+2)z}{(b+4)+{}\ddots }}}}}}}}}}}
Similarly
1
F
1
(
a
;
b
+
1
;
z
)
1
F
1
(
a
;
b
;
z
)
=
1
1
+
a
b
(
b
+
1
)
z
1
+
a
−
b
−
1
(
b
+
1
)
(
b
+
2
)
z
1
+
a
+
1
(
b
+
2
)
(
b
+
3
)
z
1
+
a
−
b
−
2
(
b
+
3
)
(
b
+
4
)
z
1
+
⋱
{\displaystyle {\frac {{}_{1}F_{1}(a;b+1;z)}{{}_{1}F_{1}(a;b;z)}}={\cfrac {1}{1+{\cfrac {{\frac {a}{b(b+1)}}z}{1+{\cfrac {{\frac {a-b-1}{(b+1)(b+2)}}z}{1+{\cfrac {{\frac {a+1}{(b+2)(b+3)}}z}{1+{\cfrac {{\frac {a-b-2}{(b+3)(b+4)}}z}{1+{}\ddots }}}}}}}}}}}
or
1
F
1
(
a
;
b
+
1
;
z
)
b
1
F
1
(
a
;
b
;
z
)
=
1
b
+
a
z
(
b
+
1
)
+
(
a
−
b
−
1
)
z
(
b
+
2
)
+
(
a
+
1
)
z
(
b
+
3
)
+
(
a
−
b
−
2
)
z
(
b
+
4
)
+
⋱
{\displaystyle {\frac {{}_{1}F_{1}(a;b+1;z)}{b{}_{1}F_{1}(a;b;z)}}={\cfrac {1}{b+{\cfrac {az}{(b+1)+{\cfrac {(a-b-1)z}{(b+2)+{\cfrac {(a+1)z}{(b+3)+{\cfrac {(a-b-2)z}{(b+4)+{}\ddots }}}}}}}}}}}
Since
1
F
1
(
0
;
b
;
z
)
=
1
{\displaystyle {}_{1}F_{1}(0;b;z)=1}
a to 0 and replacing b + 1 with b in the first continued fraction gives a simplified special case:
1
F
1
(
1
;
b
;
z
)
=
1
1
+
−
z
b
+
z
(
b
+
1
)
+
−
b
z
(
b
+
2
)
+
2
z
(
b
+
3
)
+
−
(
b
+
1
)
z
(
b
+
4
)
+
⋱
{\displaystyle {}_{1}F_{1}(1;b;z)={\cfrac {1}{1+{\cfrac {-z}{b+{\cfrac {z}{(b+1)+{\cfrac {-bz}{(b+2)+{\cfrac {2z}{(b+3)+{\cfrac {-(b+1)z}{(b+4)+{}\ddots }}}}}}}}}}}}}
The final case involves
2
F
1
(
a
,
b
;
c
;
z
)
=
1
+
a
b
c
1
!
z
+
a
(
a
+
1
)
b
(
b
+
1
)
c
(
c
+
1
)
2
!
z
2
+
a
(
a
+
1
)
(
a
+
2
)
b
(
b
+
1
)
(
b
+
2
)
c
(
c
+
1
)
(
c
+
2
)
3
!
z
3
+
⋯
.
{\displaystyle {}_{2}F_{1}(a,b;c;z)=1+{\frac {ab}{c\,1!}}z+{\frac {a(a+1)b(b+1)}{c(c+1)\,2!}}z^{2}+{\frac {a(a+1)(a+2)b(b+1)(b+2)}{c(c+1)(c+2)\,3!}}z^{3}+\cdots .\,}
Again, two identities are used alternately.
2
F
1
(
a
,
b
;
c
−
1
;
z
)
−
2
F
1
(
a
+
1
,
b
;
c
;
z
)
=
(
a
−
c
+
1
)
b
z
c
(
c
−
1
)
2
F
1
(
a
+
1
,
b
+
1
;
c
+
1
;
z
)
,
{\displaystyle \,_{2}F_{1}(a,b;c-1;z)-\,_{2}F_{1}(a+1,b;c;z)={\frac {(a-c+1)bz}{c(c-1)}}\,_{2}F_{1}(a+1,b+1;c+1;z),}
2
F
1
(
a
,
b
;
c
−
1
;
z
)
−
2
F
1
(
a
,
b
+
1
;
c
;
z
)
=
(
b
−
c
+
1
)
a
z
c
(
c
−
1
)
2
F
1
(
a
+
1
,
b
+
1
;
c
+
1
;
z
)
.
{\displaystyle \,_{2}F_{1}(a,b;c-1;z)-\,_{2}F_{1}(a,b+1;c;z)={\frac {(b-c+1)az}{c(c-1)}}\,_{2}F_{1}(a+1,b+1;c+1;z).}
These are essentially the same identity with a and b interchanged.
Let
f
0
(
z
)
=
2
F
1
(
a
,
b
;
c
;
z
)
,
{\displaystyle f_{0}(z)=\,_{2}F_{1}(a,b;c;z),}
f
1
(
z
)
=
2
F
1
(
a
+
1
,
b
;
c
+
1
;
z
)
,
{\displaystyle f_{1}(z)=\,_{2}F_{1}(a+1,b;c+1;z),}
f
2
(
z
)
=
2
F
1
(
a
+
1
,
b
+
1
;
c
+
2
;
z
)
,
{\displaystyle f_{2}(z)=\,_{2}F_{1}(a+1,b+1;c+2;z),}
f
3
(
z
)
=
2
F
1
(
a
+
2
,
b
+
1
;
c
+
3
;
z
)
,
{\displaystyle f_{3}(z)=\,_{2}F_{1}(a+2,b+1;c+3;z),}
f
4
(
z
)
=
2
F
1
(
a
+
2
,
b
+
2
;
c
+
4
;
z
)
,
{\displaystyle f_{4}(z)=\,_{2}F_{1}(a+2,b+2;c+4;z),}
etc.
This gives
f
i
−
1
−
f
i
=
k
i
z
f
i
+
1
{\displaystyle f_{i-1}-f_{i}=k_{i}zf_{i+1}}
k
1
=
(
a
−
c
)
b
c
(
c
+
1
)
,
k
2
=
(
b
−
c
−
1
)
(
a
+
1
)
(
c
+
1
)
(
c
+
2
)
,
k
3
=
(
a
−
c
−
1
)
(
b
+
1
)
(
c
+
2
)
(
c
+
3
)
,
k
4
=
(
b
−
c
−
2
)
(
a
+
2
)
(
c
+
3
)
(
c
+
4
)
{\displaystyle k_{1}={\tfrac {(a-c)b}{c(c+1)}},k_{2}={\tfrac {(b-c-1)(a+1)}{(c+1)(c+2)}},k_{3}={\tfrac {(a-c-1)(b+1)}{(c+2)(c+3)}},k_{4}={\tfrac {(b-c-2)(a+2)}{(c+3)(c+4)}}}
[ 6]
2
F
1
(
a
+
1
,
b
;
c
+
1
;
z
)
2
F
1
(
a
,
b
;
c
;
z
)
=
1
1
+
(
a
−
c
)
b
c
(
c
+
1
)
z
1
+
(
b
−
c
−
1
)
(
a
+
1
)
(
c
+
1
)
(
c
+
2
)
z
1
+
(
a
−
c
−
1
)
(
b
+
1
)
(
c
+
2
)
(
c
+
3
)
z
1
+
(
b
−
c
−
2
)
(
a
+
2
)
(
c
+
3
)
(
c
+
4
)
z
1
+
⋱
{\displaystyle {\frac {{}_{2}F_{1}(a+1,b;c+1;z)}{{}_{2}F_{1}(a,b;c;z)}}={\cfrac {1}{1+{\cfrac {{\frac {(a-c)b}{c(c+1)}}z}{1+{\cfrac {{\frac {(b-c-1)(a+1)}{(c+1)(c+2)}}z}{1+{\cfrac {{\frac {(a-c-1)(b+1)}{(c+2)(c+3)}}z}{1+{\cfrac {{\frac {(b-c-2)(a+2)}{(c+3)(c+4)}}z}{1+{}\ddots }}}}}}}}}}}
or
2
F
1
(
a
+
1
,
b
;
c
+
1
;
z
)
c
2
F
1
(
a
,
b
;
c
;
z
)
=
1
c
+
(
a
−
c
)
b
z
(
c
+
1
)
+
(
b
−
c
−
1
)
(
a
+
1
)
z
(
c
+
2
)
+
(
a
−
c
−
1
)
(
b
+
1
)
z
(
c
+
3
)
+
(
b
−
c
−
2
)
(
a
+
2
)
z
(
c
+
4
)
+
⋱
{\displaystyle {\frac {{}_{2}F_{1}(a+1,b;c+1;z)}{c{}_{2}F_{1}(a,b;c;z)}}={\cfrac {1}{c+{\cfrac {(a-c)bz}{(c+1)+{\cfrac {(b-c-1)(a+1)z}{(c+2)+{\cfrac {(a-c-1)(b+1)z}{(c+3)+{\cfrac {(b-c-2)(a+2)z}{(c+4)+{}\ddots }}}}}}}}}}}
Since
2
F
1
(
0
,
b
;
c
;
z
)
=
1
{\displaystyle {}_{2}F_{1}(0,b;c;z)=1}
a to 0 and replacing c + 1 with c gives a simplified special case of the continued fraction:
2
F
1
(
1
,
b
;
c
;
z
)
=
1
1
+
−
b
z
c
+
(
b
−
c
)
z
(
c
+
1
)
+
−
c
(
b
+
1
)
z
(
c
+
2
)
+
2
(
b
−
c
−
1
)
z
(
c
+
3
)
+
−
(
c
+
1
)
(
b
+
2
)
z
(
c
+
4
)
+
⋱
{\displaystyle {}_{2}F_{1}(1,b;c;z)={\cfrac {1}{1+{\cfrac {-bz}{c+{\cfrac {(b-c)z}{(c+1)+{\cfrac {-c(b+1)z}{(c+2)+{\cfrac {2(b-c-1)z}{(c+3)+{\cfrac {-(c+1)(b+2)z}{(c+4)+{}\ddots }}}}}}}}}}}}}
Convergence properties
edit
In this section, the cases where one or more of the parameters is a negative integer are excluded, since in these cases either the hypergeometric series are undefined or that they are polynomials so the continued fraction terminates. Other trivial exceptions are excluded as well.
In the cases
0
F
1
{\displaystyle {}_{0}F_{1}}
1
F
1
{\displaystyle {}_{1}F_{1}}
meromorphic function . The continued fractions on the right hand side will converge uniformly on any closed and bounded set that contains no poles of this function.[ 7]
In the case
2
F
1
{\displaystyle {}_{2}F_{1}}
Outside the circle, the continued fraction represents the analytic continuation of the function to the complex plane with the positive real axis, from +1 to the point at infinity removed. In most cases +1 is a branch point and the line from +1 to positive infinity is a branch cut for this function. The continued fraction converges to a meromorphic function on this domain, and it converges uniformly on any closed and bounded subset of this domain that does not contain any poles.[ 8]
We have
cosh
(
z
)
=
0
F
1
(
1
2
;
z
2
4
)
,
{\displaystyle \cosh(z)=\,_{0}F_{1}({\tfrac {1}{2}};{\tfrac {z^{2}}{4}}),}
sinh
(
z
)
=
z
0
F
1
(
3
2
;
z
2
4
)
,
{\displaystyle \sinh(z)=z\,_{0}F_{1}({\tfrac {3}{2}};{\tfrac {z^{2}}{4}}),}
so
tanh
(
z
)
=
z
0
F
1
(
3
2
;
z
2
4
)
0
F
1
(
1
2
;
z
2
4
)
=
z
/
2
1
2
+
z
2
4
3
2
+
z
2
4
5
2
+
z
2
4
7
2
+
⋱
=
z
1
+
z
2
3
+
z
2
5
+
z
2
7
+
⋱
.
{\displaystyle \tanh(z)={\frac {z\,_{0}F_{1}({\tfrac {3}{2}};{\tfrac {z^{2}}{4}})}{\,_{0}F_{1}({\tfrac {1}{2}};{\tfrac {z^{2}}{4}})}}={\cfrac {z/2}{{\tfrac {1}{2}}+{\cfrac {\tfrac {z^{2}}{4}}{{\tfrac {3}{2}}+{\cfrac {\tfrac {z^{2}}{4}}{{\tfrac {5}{2}}+{\cfrac {\tfrac {z^{2}}{4}}{{\tfrac {7}{2}}+{}\ddots }}}}}}}}={\cfrac {z}{1+{\cfrac {z^{2}}{3+{\cfrac {z^{2}}{5+{\cfrac {z^{2}}{7+{}\ddots }}}}}}}}.}
This particular expansion is known as Lambert's continued fraction and dates back to 1768.[ 9]
It easily follows that
tan
(
z
)
=
z
1
−
z
2
3
−
z
2
5
−
z
2
7
−
⋱
.
{\displaystyle \tan(z)={\cfrac {z}{1-{\cfrac {z^{2}}{3-{\cfrac {z^{2}}{5-{\cfrac {z^{2}}{7-{}\ddots }}}}}}}}.}
The expansion of tanh can be used to prove that e n n (which is alas not enough to prove that e is transcendental ). The expansion of tan was used by both Lambert and Legendre to prove that π is irrational .
The Bessel function
J
ν
{\displaystyle J_{\nu }}
J
ν
(
z
)
=
(
1
2
z
)
ν
Γ
(
ν
+
1
)
0
F
1
(
ν
+
1
;
−
z
2
4
)
,
{\displaystyle J_{\nu }(z)={\frac {({\tfrac {1}{2}}z)^{\nu }}{\Gamma (\nu +1)}}\,_{0}F_{1}(\nu +1;-{\frac {z^{2}}{4}}),}
from which it follows
J
ν
(
z
)
J
ν
−
1
(
z
)
=
z
2
ν
−
z
2
2
(
ν
+
1
)
−
z
2
2
(
ν
+
2
)
−
z
2
2
(
ν
+
3
)
−
⋱
.
{\displaystyle {\frac {J_{\nu }(z)}{J_{\nu -1}(z)}}={\cfrac {z}{2\nu -{\cfrac {z^{2}}{2(\nu +1)-{\cfrac {z^{2}}{2(\nu +2)-{\cfrac {z^{2}}{2(\nu +3)-{}\ddots }}}}}}}}.}
These formulas are also valid for every complex z .
Since
e
z
=
1
F
1
(
1
;
1
;
z
)
{\displaystyle e^{z}={}_{1}F_{1}(1;1;z)}
1
/
e
z
=
e
−
z
{\displaystyle 1/e^{z}=e^{-z}}
e
z
=
1
1
+
−
z
1
+
z
2
+
−
z
3
+
2
z
4
+
−
2
z
5
+
⋱
{\displaystyle e^{z}={\cfrac {1}{1+{\cfrac {-z}{1+{\cfrac {z}{2+{\cfrac {-z}{3+{\cfrac {2z}{4+{\cfrac {-2z}{5+{}\ddots }}}}}}}}}}}}}
e
z
=
1
+
z
1
+
−
z
2
+
z
3
+
−
2
z
4
+
2
z
5
+
⋱
.
{\displaystyle e^{z}=1+{\cfrac {z}{1+{\cfrac {-z}{2+{\cfrac {z}{3+{\cfrac {-2z}{4+{\cfrac {2z}{5+{}\ddots }}}}}}}}}}.}
With some manipulation, this can be used to prove the simple continued fraction representation of
e
e
=
2
+
1
1
+
1
2
+
1
1
+
1
1
+
1
4
+
⋱
{\displaystyle e=2+{\cfrac {1}{1+{\cfrac {1}{2+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{4+{}\ddots }}}}}}}}}}}
The error function erf (z ), given by
erf
(
z
)
=
2
π
∫
0
z
e
−
t
2
d
t
,
{\displaystyle \operatorname {erf} (z)={\frac {2}{\sqrt {\pi }}}\int _{0}^{z}e^{-t^{2}}\,dt,}
can also be computed in terms of Kummer's hypergeometric function:
erf
(
z
)
=
2
z
π
e
−
z
2
1
F
1
(
1
;
3
2
;
z
2
)
.
{\displaystyle \operatorname {erf} (z)={\frac {2z}{\sqrt {\pi }}}e^{-z^{2}}\,_{1}F_{1}(1;{\scriptstyle {\frac {3}{2}}};z^{2}).}
By applying the continued fraction of Gauss, a useful expansion valid for every complex number z can be obtained:[ 10]
π
2
e
z
2
erf
(
z
)
=
z
1
−
z
2
3
2
+
z
2
5
2
−
3
2
z
2
7
2
+
2
z
2
9
2
−
5
2
z
2
11
2
+
3
z
2
13
2
−
7
2
z
2
15
2
+
−
⋱
.
{\displaystyle {\frac {\sqrt {\pi }}{2}}e^{z^{2}}\operatorname {erf} (z)={\cfrac {z}{1-{\cfrac {z^{2}}{{\frac {3}{2}}+{\cfrac {z^{2}}{{\frac {5}{2}}-{\cfrac {{\frac {3}{2}}z^{2}}{{\frac {7}{2}}+{\cfrac {2z^{2}}{{\frac {9}{2}}-{\cfrac {{\frac {5}{2}}z^{2}}{{\frac {11}{2}}+{\cfrac {3z^{2}}{{\frac {13}{2}}-{\cfrac {{\frac {7}{2}}z^{2}}{{\frac {15}{2}}+-\ddots }}}}}}}}}}}}}}}}.}
A similar argument can be made to derive continued fraction expansions for the Fresnel integrals , for the Dawson function , and for the incomplete gamma function . A simpler version of the argument yields two useful continued fraction expansions of the exponential function .[ 11]
From
(
1
−
z
)
−
b
=
1
F
0
(
b
;
;
z
)
=
2
F
1
(
1
,
b
;
1
;
z
)
,
{\displaystyle (1-z)^{-b}={}_{1}F_{0}(b;;z)=\,_{2}F_{1}(1,b;1;z),}
(
1
−
z
)
−
b
=
1
1
+
−
b
z
1
+
(
b
−
1
)
z
2
+
−
(
b
+
1
)
z
3
+
2
(
b
−
2
)
z
4
+
⋱
{\displaystyle (1-z)^{-b}={\cfrac {1}{1+{\cfrac {-bz}{1+{\cfrac {(b-1)z}{2+{\cfrac {-(b+1)z}{3+{\cfrac {2(b-2)z}{4+{}\ddots }}}}}}}}}}}
It is easily shown[ 12] arctan z in a neighborhood of zero is given by
arctan
z
=
z
F
(
1
2
,
1
;
3
2
;
−
z
2
)
.
{\displaystyle \arctan z=zF({\scriptstyle {\frac {1}{2}}},1;{\scriptstyle {\frac {3}{2}}};-z^{2}).}
The continued fraction of Gauss can be applied to this identity, yielding the expansion
arctan
z
=
z
1
+
(
1
z
)
2
3
+
(
2
z
)
2
5
+
(
3
z
)
2
7
+
(
4
z
)
2
9
+
⋱
,
{\displaystyle \arctan z={\cfrac {z}{1+{\cfrac {(1z)^{2}}{3+{\cfrac {(2z)^{2}}{5+{\cfrac {(3z)^{2}}{7+{\cfrac {(4z)^{2}}{9+\ddots }}}}}}}}}},}
which converges to the principal branch of the inverse tangent function on the cut complex plane, with the cut extending along the imaginary axis from i to the point at infinity, and from −i to the point at infinity.[ 13]
This particular continued fraction converges fairly quickly when z = 1, giving the value π/4 to seven decimal places by the ninth convergent. The corresponding series
π
4
=
1
1
+
1
2
2
+
3
2
2
+
5
2
2
+
⋱
=
1
−
1
3
+
1
5
−
1
7
±
⋯
{\displaystyle {\frac {\pi }{4}}={\cfrac {1}{1+{\cfrac {1^{2}}{2+{\cfrac {3^{2}}{2+{\cfrac {5^{2}}{2+\ddots }}}}}}}}=1-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}\pm \cdots }
converges much more slowly, with more than a million terms needed to yield seven decimal places of accuracy.[ 14]
Variations of this argument can be used to produce continued fraction expansions for the natural logarithm , the arcsin function , and the generalized binomial series .
^ Jones & Thron (1980) p. 5
^ C. F. Gauss (1813), Werke , vol. 3
^ B. Riemann (1863), "Sullo svolgimento del quoziente di due serie ipergeometriche in frazione continua infinita" in Werke . pp. 400–406. (Posthumous fragment).
^ L. W. Thomé (1867), "Über die Kettenbruchentwicklung des Gauß'schen Quotienten ...," Jour. für Math. vol. 67 pp. 299–309.
^ E. B. Van Vleck (1901), "On the convergence of the continued fraction of Gauss and other continued fractions." Annals of Mathematics , vol. 3 pp. 1–18.
^ Frank, E (1956). "A new class of continued fraction expansions for the ratios of hypergeometric functions". Trans. Am. Math. Soc . 81 (2): 453–476. JSTOR 1992927 . MR 0076937 . ^ Jones & Thron (1980) p. 206
^ Wall, 1973 (p. 339)
^ Wall (1973) p. 349.
^ Jones & Thron (1980) p. 208.
^ See the example in the article Padé table for the expansions of ez as continued fractions of Gauss.
^ ProofWiki ^ Wall (1973) p. 343. Notice that i and −i are branch points for the inverse tangent function.
^ Jones & Thron (1980) p. 202.
