Continued fractions for confluent hypergeometric functions
Today we show the proof of this nice continued fraction proposed by @nntaleb xex−1=1−x2+x3−2x4+2x5−3x6+⋯
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+n−1)
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+2n−1)(c+2n)n=1,2,3,...
a2n+1=−c−b+n(c+2n)(c+2n+1)n=0,1,2,3,...
Then
1+K∞n=1(anz1)=1F1(b;c;z)1F1(b+1;c+1;z)∀z∈C
Corollary If we let b→0 and making the transformation c+1↦c
1−1cz1+1c(c+1)z1−c(c+1)(c+2)z1+2(c+2)(c+3)z1−c+1(c+3)(c+4)z1+3(c+4)(c+5)z1−⋯=11F1(1;c;z)
⟹1F1(1;c;z)=11−1cz1+1c(c+1)z1−c(c+1)(c+2)z1+2(c+2)(c+3)z1−c+1(c+3)(c+4)z1+3(c+4)(c+5)z1−⋯=11−zc+zc+1−czc+1+2zc+3−(c+1)zc+4+3zc+5−⋯
Applying (∗) to ex:
ex=1F1(1;1;x)=11−x1+x2−x3+2x4−2x5+3z6−⋯
Making the transformation x↦−x:
⟹e−x=1ex=11+x1−x2+x3−2x4+2x5−3z6+⋯
⟹ex=1+x1−x2+x3−2x4+2x5−3z6+⋯
⟹ex−1=x1−x2+x3−2x4+2x5−3z6+⋯
Finally, taking the reciprocal:
⟹xex−1=1−x2+x3−2x4+2x5−3z6+⋯
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