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