Gauss's continued fraction

In complex analysis, Gauss's continued fraction is a particular class of continued fractions derived from hypergeometric functions. It was one of the first analytic continued fractions known to mathematics, and it can be used to represent several important elementary functions, as well as some of the more complicated transcendental functions.

History

Lambert published several examples of continued fractions in this form in 1768, and both Euler and Lagrange investigated similar constructions,^[1] but it was 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] and L.W. Thomé^[4] obtained partial results, but the final word on the region in which this continued fraction converges was not given until 1901, by Edward Burr Van Vleck.^[5]

Derivation

Let $f_{0},f_{1},f_{2},\dots$ be a sequence of analytic functions so that

f_{{i-1}}-f_{i}=k_{i}\,z\,f_{{i+1}}

for all $i>0$ , where each $k_{i}$ is a constant.

Then

{\frac {f_{{i-1}}}{f_{i}}}=1+k_{i}z{\frac {f_{{i+1}}}{{f_{i}}}},\,

and so

{\frac {f_{i}}{f_{{i-1}}}}={\frac {1}{1+k_{i}z{\frac {f_{{i+1}}}{{f_{i}}}}}}

Setting $g_{i}=f_{i}/f_{{i-1}}$ ,

g_{i}={\frac {1}{1+k_{i}zg_{{i+1}}}}

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}}}}}}}=\dots \

Repeating this ad infinitum produces the continued fraction expression

{\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 $f_{i}$ are hypergeometric functions of the form ${}_{0}F_{1}$ , ${}_{1}F_{1}$ , and ${}_{2}F_{1}$ , and the equations $f_{{i-1}}-f_{i}=k_{i}zf_{{i+1}}$ arise as identities between functions where the parameters differ by integer amounts. These identities can be proven in several ways, for example by expanding out the series and comparing coefficients, or by taking the derivative in several ways and eliminating it from the equations generated.

The series ₀F₁

The simplest case involves

\,_0F_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

\,_0F_1(a-1;z)-\,_0F_1(a;z) = \frac{z}{a(a-1)}\,_0F_1(a+1;z)

we may take

f_i = {}_0F_1(a+i;z),\,k_i = \tfrac{1}{(a+i)(a+i-1)}

giving

{\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 }}}}}}}}

{\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 series ₁F₁

The next case involves

{}_{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}+\dots

for which the two identities

\,_{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)={\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)

f_{1}(z)=\,_{1}F_{1}(a+1;b+1;z)

f_{2}(z)=\,_{1}F_{1}(a+1;b+2;z)

f_{3}(z)=\,_{1}F_{1}(a+2;b+3;z)

f_{4}(z)=\,_{1}F_{1}(a+2;b+4;z)

etc.

This gives $f_{{i-1}}-f_{i}=k_{i}zf_{{i+1}}$ where $k_{1}={\tfrac {a-b}{b(b+1)}},k_{2}={\tfrac {a+1}{(b+1)(b+2)}},k_{3}={\tfrac {a-b-1}{(b+2)(b+3)}},k_{4}={\tfrac {a+2}{(b+3)(b+4)}}$ , producing

{\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 }}}}}}}}}}

{\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

{\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 }}}}}}}}}}

{\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 ${}_{1}F_{1}(0;b;z)=1$ , setting 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)={\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 series ₂F₁

The final case involves

{}_{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}+\dots \,

Again, two identities are used alternately.

\,_{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)={\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)

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)

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)

etc.

This gives $f_{{i-1}}-f_{i}=k_{i}zf_{{i+1}}$ where $k_{1}={\tfrac {(a-c)b}{c(c+1)}},k_{2}={\tfrac {(b-c-1)(a+1)}{(c+1)(c+2)}},k_{3}={\tfrac {(a-c-1)(b+1)}{(c+2)(c+3)}},k_{4}={\tfrac {(b-c-2)(a+2)}{(c+3)(c+4)}}$ , producing

{\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 }}}}}}}}}}

{\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 ${}_{2}F_{1}(0,b;c;z)=1$ , setting 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)={\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

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 ${}_{0}F_{1}$ and ${}_{1}F_{1}$ , the series converge everywhere so the fraction on the left hand side is a 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.^[6]

In the case ${}_{2}F_{1}$ , the radius of convergence of the series is 1 and the fraction on the left hand side is a meromorphic function within this circle. The continued fractions on the right hand side will converge to the function everywhere inside this circle.

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.^[7]