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Re: Hypergeometric1F1 polynomial

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  • Subject: [mg91461] Re: Hypergeometric1F1 polynomial
  • From: "Alec Mihailovs" <alec at>
  • Date: Fri, 22 Aug 2008 03:14:16 -0400 (EDT)
  • References: <g8je5u$a4n$> <>

From: "Jean-Marc Gulliet" <jeanmarc.gulliet at>
> 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[1]:= Assuming[Element[n, Integers] && n > 0,
>      FullSimplify[
>       Sum[Binomial[n, k]/Binomial[2 n, k]/k! (2 x)^k, {k, 0, n}]]]
>     Out[1]= E^x Hypergeometric0F1[1/2 - n, x^2/4]

But that is the wrong answer as well. The sum is a polynomial of x of degree 
n, while Out[1] is not. For example,

In[2]:= % /. n -> 1

Out[2]= -(1/2) E^x x (-((2 Cosh[x])/x) + 2 Sinh[x])

while it should be 1+x.

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

That is a bug. They are not equal for positive integer n. One is a 
polynomial of x, and another one is not.


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