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Re: Mathematica and infinite series

  • To: mathgroup at smc.vnet.net
  • Subject: [mg113200] Re: Mathematica and infinite series
  • From: Sam Takoy <sam.takoy at yahoo.com>
  • Date: Mon, 18 Oct 2010 05:48:09 -0400 (EDT)

Hi,

Thanks for the response!

I don't have Mathematica on the computer on which I'm typing this email, so I'm 
not able to reformulate my question reliably.

I was hoping to come up with a series that does not have a closed form 
analytical expression. I was hoping that Mathematica would be able to treat 
SeriesCoefficient and an inverse to an infinite Sum. How about (hoping it's 
"safe" from Mathematica's analytical ability):

f[x_] := Sum[BesselJZero[0, n^2] Sin[n^2] Log[Sin[Cos[n]]]Log[n]/(n^2 
Factorial[n]) x^n, {n, 1, Infinity}]
Assuming[n > 0, SeriesCoefficient[f[x], {x, 0, n}]]


Thanks again,

Sam




________________________________
From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
To: Sam Takoy <sam.takoy at yahoo.com>
Sent: Sun, October 17, 2010 12:05:30 PM
Subject: [mg113200] Re: [mg113185] Mathematica and infinite series

The problem is that Mathematica seems to go crazy when asked to evaluate:

Sum[(Log[n]/(n^2*n!))*x^n, {n, 1, Infinity}]

E^x*Derivative[1, 0][BellB][-2, x]

The answer is given in terms of the derivative of the function BellB[n,x] with 
respect to the first variable at {-2,x}. But, and this is weird, since Bell[n,x] 
is the n-th Bell polynomial so n must be a non-negative integer:

In[4]:= BellB[-1,2]
During evaluation of In[4]:= BellB::intnm: Non-negative machine-size integer 
expected at position 1 in Subscript[B, -1](2). >>


See also here : http://mathworld.wolfram.com/BellPolynomial.html

Really weird. 

Andrzej Kozlowski


On 17 Oct 2010, at 12:06, Sam Takoy wrote:

> Hi,
> 
> I am about to embark on a project that operates heavily in infinite 
> series, so I started figuring out Mathematica's basis capabilities. I 
> found them very impressive, but I came across this:
> 
> 
> f[x_] := Sum[Log[n]/(n^2 Factorial[n]) x^n, {n, 1, Infinity}]
> Assuming[n > 0, SeriesCoefficient[f[x], {x, 0, 4}]]
> 
> 
> Answer:
> 
> SeriesCoefficient[\!\(
> \*UnderoverscriptBox[\(\[Sum]\), \(n = 1\), \(\[Infinity]\)]
> \*FractionBox[\(
> \*SuperscriptBox[\(x\), \(n\)]\ Log[n]\), \(
> \*SuperscriptBox[\(n\), \(2\)]\ \(n!\)\)]\), {x, 0, 4}]
> 
> 
> Why doesn't Mathematica produce Log[n]/(n^2 Factorial[n]) as the answer?
> 
> Thanks!
> 
> Sam
> 


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