Re: Hypergeometric1F1 polynomial
- To: mathgroup at smc.vnet.net
- Subject: [mg91447] Re: Hypergeometric1F1 polynomial
- From: "Jean-Marc Gulliet" <jeanmarc.gulliet at gmail.com>
- Date: Fri, 22 Aug 2008 03:11:42 -0400 (EDT)
- References: <g8je5u$a4n$1@smc.vnet.net> <48ADCC77.9070400@gmail.com>
On Thu, Aug 21, 2008 at 11:18 PM, Alec Mihailovs <alec at mihailovs.com> wrote:
> From: "Jean-Marc Gulliet" <jeanmarc.gulliet at gmail.com>
>>
>> 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.
The polynomial form you are expecting (see In[1]) can be obtained by
taking the series expansion about x == 0 to the order n (see In[2]).
In[1]:= Table[Sum[Binomial[n, k]/Binomial[2 n, k]/k! (2 x)^k, {k, 0,
n}], {n, 0,
5}] // TableForm
Out[1]//TableForm=
1
1 + x
2
x
1 + x + --
3
2 3
2 x x
1 + x + ---- + --
5 15
2 3 4
3 x 2 x x
1 + x + ---- + ---- + ---
7 21 105
2 3 4 5
4 x x x x
1 + x + ---- + -- + -- + ---
9 9 63 945
In[2]:= Table[Series[Hypergeometric1F1[-n, -2 n, 2 x], {x, 0, n}] //
Normal, {n, 0,
5}] // TableForm
Out[2]//TableForm=
1
1 + x
2
x
1 + x + --
3
2 3
2 x x
1 + x + ---- + --
5 15
2 3 4
3 x 2 x x
1 + x + ---- + ---- + ---
7 21 105
2 3 4 5
4 x x x x
1 + x + ---- + -- + -- + ---
9 9 63 945
Regards,
--
Jean-Marc