Re: Hypergeometric1F1 polynomial

*To*: mathgroup at smc.vnet.net*Subject*: [mg91470] Re: Hypergeometric1F1 polynomial*From*: "Alec Mihailovs" <alec at mihailovs.com>*Date*: Sat, 23 Aug 2008 01:40:24 -0400 (EDT)*References*: <g8je5u$a4n$1@smc.vnet.net> <48ADCC77.9070400@gmail.com> <DA73101988B04F9CAAD09D9816E75229@AlecPC> <22d35c5a0808212132x2413bd58pd700a1b5cdac9ae4@mail.gmail.com> <2B30ED560CEC44A8AD5D6CE21233E7C6@AlecPC> <22d35c5a0808220025g24090120pc9a35e99332801c6@mail.gmail.com>

From: "Jean-Marc Gulliet" <jeanmarc.gulliet at gmail.com> > On Fri, Aug 22, 2008 at 7:13 AM, Alec Mihailovs <alec at mihailovs.com> > wrote: >> The problem is that the answers given by Mathematica to the Sum problem, >> are >> not the same - they are not polynomials, with the series expansion, or >> without. > > Hum, with series expansion they are (at least on my system). For instance, > > In[1]:= s = Sum[Binomial[n, k]/Binomial[2 n, k]/k! (2 x)^k, {k, 0, n}] > > Out[1]= > > -(1/2) - n x 1/2 + n 1 1 > 2 E x BesselI[- (-1 - 2 n), x] Gamma[- - n] > 2 2 > > In[2]:= Table[Series[s, {x, 0, n}] // Normal, {n, 0, 5}] // TableForm That just tells that the beginning of the series is the same. But the rest should be subtracted to get the correct answer - that's generally how the answer with BesselK appears - as a result of subtracting of 2 expressions with BesselI. > In[3]:= FullSimplify[s] > Table[Series[%, {x, 0, n}] // Normal, {n, 0, 5}] // TableForm The same here - if you took more than n terms from the series, you would see the difference. 1 is not equal to E^(2x) even if the first term of Taylor series is the same :) > In[7]:= Table[FullSimplify[s] == Hypergeometric1F1[-n, -2 n, 2 x], {n, > 0, 5}, {x, 1, > 5}] > > Out[7]= {{False, False, False, False, False}, {False, False, False, False, > False}, {False, False, False, False, False}, {False, False, False, False, > False}, {False, False, False, False, False}, {False, False, False, False, > False}} That, certainly, is correct - the answer given by Mathematica is not equal to the correct answer (as I said in the original post). Alec