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Friday, January 28, 2022

Continued fractions II

Kummer's transformation

Continued fractions for confluent hypergeometric functions


Today we show the proof of this nice continued fraction proposed by @nntaleb xex1=1x2+x32x4+2x53x6+
The proof will relly on the continued fractions theorem for confluent hypergeometric functions

Proof

Note that ex=n=0xnn!=n=0(1)n(1)nxnn!=1F1(1;1;x)
where (x)n=x(x+1)(x+2)(x+n1)
is the Pochhhammer polynomial or rising factorial of degree n and 1F1(a;b;z) is the confluent hypergeometric function, also denoted by M(a;b;z). There is a continued fractions theorem for confluent hypergeometric functions:

Thorem Consider the function:

1F1(b;c;z)=n=0(b)n(c)nznn!
Let {an} be a sequence of complex numbers defined by a2n=b+n(c+2n1)(c+2n)n=1,2,3,...
a2n+1=cb+n(c+2n)(c+2n+1)n=0,1,2,3,...
Then 1+Kn=1(anz1)=1F1(b;c;z)1F1(b+1;c+1;z)zC


Corollary If we let b0 and making the transformation c+1c

11cz1+1c(c+1)z1c(c+1)(c+2)z1+2(c+2)(c+3)z1c+1(c+3)(c+4)z1+3(c+4)(c+5)z1=11F1(1;c;z)
1F1(1;c;z)=111cz1+1c(c+1)z1c(c+1)(c+2)z1+2(c+2)(c+3)z1c+1(c+3)(c+4)z1+3(c+4)(c+5)z1=11zc+zc+1czc+1+2zc+3(c+1)zc+4+3zc+5
Applying () to ex: ex=1F1(1;1;x)=11x1+x2x3+2x42x5+3z6
Making the transformation xx: ex=1ex=11+x1x2+x32x4+2x53z6+
ex=1+x1x2+x32x4+2x53z6+
ex1=x1x2+x32x4+2x53z6+
Finally, taking the reciprocal: xex1=1x2+x32x4+2x53z6+

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