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Wednesday, December 8, 2021

Confluent hypergeometric function

Fun series

Singular series posted by @infseriesbot


Today we evaluate this series posted by @infseriesbot (n=0(2n)!n!3xn)2=e4xn=0(2n)!n!4x2n The proof will rely on some properties of the confluent hypergeometric function and the binomial coefficient

Proof:

We start with the left hand side. Recall that the Pocchhammer symbol or rising factorial can be expressed as quotient of two gamma functions (x)n=Γ(n+x)Γ(x) Hence (2n)!=Γ(2n+1)=(1)2n Pochhammer symbols also satisfy the duplication formula: (x)2n=4n(x2)n(1+x2)n Hence (2n)!=4n(12)n(1)n Note also that (1)n=n!. Hence, n=0(2n)!n!3xn=n=04n(12)n(1)n(1)n(1)nxnn!=n=0(12)n(1)n(4x)nn!=1F1(12;1;4x) The right hand side is the confluent hypergometric function 1F1(a;b;x). This function is also denoted with M(a;b;x):

The confluent hypergometric function satisfies the Kummer's second theorem or second transformation: 1F1(a,2a,x)=ex/20F1(;a+12;x216) Therefore n=0(2n)!n!3xn=1F1(12;1;4x)=e2x0F1(;1;x2) From the definition of 0F1: e2x0F1(;1;x2)=e2xn=0x2nn!2 Hence n=0(2n)!n!3xn=e2xn=0x2nn!2 Squaring both sides (n=0(2n)!n!3xn)2=e4x(n=0x2nn!2)2 We can evaluate the right hand side with the Cauchy product: (n=0x2nn!2)2=n=0nj=01j!2(nj)!2x2n=n=01n!2nj=0n!2j!2(nj)!2x2n=n=0x2nn!2nj=0(nj)2 From the properties of the binomial coefficient we know that: Jj=0(nj)2=(2nn)Jn Therefore (n=0x2nn!2)2=n=0x2nn!2nj=0(nj)2=n=0x2nn!2(2nn)=n=0(2n)!n!4x2n Hence from (1),(2),(3) and (4) we can conclude: (n=0(2n)!n!3xn)2=e4xn=0(2n)!n!4x2n

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