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Wednesday, January 26, 2022

Confluent hypergeometric function II

Kummer's transformation

Kummer's transformation for 1F1


Here is the proof of this transformation posted by @infseriesbot n=0zn(a)n+1=ezn=0(z)n(a+n)


Proof:

Recall the definition of the Pochhammer polynomial (rising factorial) (a)n=a(a+1)(a+n1)
Hence S=n=0zn(a)n+1=n=0zna(a+1)(a+n)=n=0zna[(a+1)(a+n)]=n=0zna(a+1)n=1an=0(1)n(a+1)nzn(1)n=1an=0(1)n(a+1)nznn!=1a1F1(1,a+1,z)
1F1(a,b,z) is the confluent hypergeometric function, also denoted M(a,b,z). This function satisfies the Kummer’s Transformation: 1F1(a,b,z)=ez1F1(ba,b,z)
Therefore 1a1F1(1,a+1,z)=eza1F1(a,a+1,z)=ezan=0(a)n(a+1)n(z)n=ezan=0a(a+1)(a+n1)(a+1)(a+2)(a+n1)(a+n)(z)n=ezan=0a(a+1)(a+n1)(a+1)(a+2)(a+n1)(a+n)(z)n=ezan=0a(a+n)(z)n=ezn=0(z)n(a+n)
Hence n=0zn(a)n+1=ezn=0(z)n(a+n)

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