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Bug in SeriesCoefficient in 7.0

  • To: mathgroup at smc.vnet.net
  • Subject: [mg97991] Bug in SeriesCoefficient in 7.0
  • From: "Maxim A. Dubinnyi" <maxim at nmr.ru>
  • Date: Fri, 27 Mar 2009 05:36:10 -0500 (EST)

There is an expression to obtain the order Nk[n] of the hyperdeterminant 
of 2^n matrix
(http://en.wikipedia.org/wiki/Hyperdeterminant):

Sum[Nk[n] z^n/n!, {n, 0, Infinity}] == Exp[-2z]/(1-z)^2

I'm trying to expand right hand side of this equation to obtain Nk[n] 
values.
First, look for several low-order coefficients:

In[1] = Ser = Series[Exp[-2 z]/(1 - z)^2, {z, 0, 9}]
Out[1] = 1 + z^2 + (2 z^3)/3 + z^4 + (16 z^5)/15 + (11 z^6)/9 + (
 142 z^7)/105 + (67 z^8)/45 + (4604 z^9)/2835 + O[z^10]

Extract Nk[n]:

In[2] = Table[n! Coefficient[Ser, z, n], {n, 0, 9}]
Out[2] = {1, 0, 2, 4, 24, 128, 880, 6816, 60032, 589312}

This is correct. Then I'm trying to get general expression for Nk[n]:

In[3] = SeriesCoefficient[Exp[-2 z]/(1 - z)^2, {z, 0, n}]
Out[3] = Piecewise[{{(3 + n)/E^2, n >= 0}}, 0]

THIS IS ABSOLUTELY WRONG ANSWER.

But the correct answer is given by Mathematica 6.0.1:

In[1] (Math 6.0.1) = Nk6[n_] = SeriesCoefficient[Exp[-2 z]/(1 - z)^2, 
{z, 0, n}]
Out[1] (Math 6.0.1) =  ((2 + n) ((-1)^n 2^(2 + n) E^2 + (3 + n) Gamma[2 
+ n, -2]))/(E^2 Gamma[3 + n])

In[2] (Math 6.0.1) = Table[n! Nk6[n], {n, 0, 9}] // FunctionExpand
Out[2] (Math 6.0.1) = {1, 0, 2, 4, 24, 128, 880, 6816, 60032, 589312}

This is again the correct answer.

Maxim A. Dubinnyi,
Shemiakin-Ovchinnikov Institue of Bioorganic Chemistry,
Moscow, Russia


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