       Re: Hypergeometric1F1 polynomial

• To: mathgroup at smc.vnet.net
• Subject: [mg91460] Re: Hypergeometric1F1 polynomial
• From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
• Date: Fri, 22 Aug 2008 03:14:04 -0400 (EDT)
• Organization: The Open University, Milton Keynes, UK
• References: <g8je5u\$a4n\$1@smc.vnet.net>

```Alec Mihailovs wrote:

> Mathematica gives the wrong answer to the following sum,
>
> In:= Sum[Binomial[n, k]/Binomial[2 n, k]/k! (2 x)^k, {k, 0, n}]
>
> Out= 2^(-(1/2) - n) E^x x^(1/2 + n)
>   BesselI[1/2 (-1 - 2 n), x] Gamma[1/2 - n]
>
> The correct answer is 1 for n=0 and Hypergeometric1F1[-n, -2 n, 2 x] for
> integer n>0, which would be equal to the expression given by Mathematica if
> n was not a positive integer.
>
> Another form of the correct answer is
>
> (2 x)^(n+1/2) E^x BesselK[n+1/2,x] n!/(2 n)!/Sqrt[Pi]
>
> Is there a way to apply some assumptions to get the correct answer?

Alec,

One can pass assumptions thanks to the function *Assuming[]* or the
option *Assumptions*, usually in combination with functions such as
Simplify or FullSimplify (when special functions are involved). For
instance,

In:= Assuming[Element[n, Integers] && n > 0,
FullSimplify[
Sum[Binomial[n, k]/Binomial[2 n, k]/k! (2 x)^k, {k, 0, n}]]]

Out= E^x Hypergeometric0F1[1/2 - n, x^2/4]

Note that the original result you got is equivalent for all n, indeed,
to the hypergeometric function you claim to be the correct solution.

In:= s =
Sum[Binomial[n, k]/Binomial[2 n, k]/k! (2 x)^k, {k, 0, n}]

Out= 2^(-(1/2) - n) E^x x^(1/2 + n)
BesselI[1/2 (-1 - 2 n), x] Gamma[1/2 - n]

In:= FullSimplify[s]

Out= E^x Hypergeometric0F1[1/2 - n, x^2/4]

In:= FullSimplify[s == Hypergeometric1F1[-n, -2 n, 2 x]]

Out= True

In:= s /. n -> 0

Out= E^x Cosh[x]

In:= % // TrigToExp

Out= 1/2 + E^(2 x)/2

The above result, however, does not match the following:

In:= With[{n = 0},
Sum[Binomial[n, k]/Binomial[2 n, k]/k! (2 x)^k, {k, 0, n}]]

Out= 1

(So the sum is now equal to one for n == 0, as you claimed.)

The *With[]* construct rewrite the sum as

Sum[Binomial[0, k]/Binomial[2 0, k]/k! (2 x)^k, {k, 0, 0}]]

*before* evaluating it.

Note that Mathematica fails in a weird way ( function == 0 ) checking
the equivalence of

In:= FullSimplify[(2 x)^(n + 1/2) E^x BesselK[n + 1/2,
x] n!/(2 n)!/Sqrt[Pi] == Hypergeometric1F1[-n, -2 n, 2 x]]

Out= (2^-n E^x x^(1/2 + n) BesselI[1/2 + n, x] Sec[n \[Pi]])/
Gamma[1/2 + n] == 0

Best regards,
-- Jean-Marc

```

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