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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
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