Help with a generating function and differential equation I have a generating function that I'm

osteoblogda 2022-01-17 Answered
Help with a generating function and differential equation
I have a generating function that Im
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Jimmy Macias
Answered 2022-01-18 Author has 30 answers
The general form of your sum can be expressed succinctly as
(c)k(x)k
where (a)k is a Pochhammer symbol.
We recognize at once that this is a hypergeometric series; specifically, it is a 2F0:
2F0(c,1:x)
which can be rewritten as a Tricomi confluent hypergeometric function:
1xU(1,c+2,1x)
which can be re-expressed as an incomplete gamma function:
xcexp(1x)(c+1,1x)
As for the differential equation, it can be obtained from this formula to give:
x2d2ydx2(c21x)xdydxcy=0
where the solution of interest satisfies the initial conditions y(0)=1 and y(0)=c

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Travis Hicks
Answered 2022-01-19 Author has 29 answers
If you know about the incomplete Gamma function then the solution is immediate
Γ(s,x)=(s1)Γ(s1,x)+xs1ex
hence exΓ(s+1,x)=k=0ss!k!xk for sN
Your method of summing the lower-factorical coefficient recurrence c(k+1)=(ck)c(k) will yield a hypergeometric differential equation with the above solution - presuming that you know how to solve such hypergeometric differential equations. But that's overkill compared to the above.

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alenahelenash
Answered 2022-01-24 Author has 368 answers
I found a differential equation which gives the correct result under Mathematica 7.0.1. The differential equation is the same one I derived in the question (see my comment on adding x): A(x)1x=cA(x)dA(x)dx The result it gives as the solution (besides the series) is: e1/xxc(C1xc1Ec(1x)) where Ec(z) is the Exponential Integral Function, for which the following equation holds: xc1Ec(b)=bcxcΓ[1c,b] and b=1x. C1 is the constant of integration.

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