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MathGroup Archive 2006

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Re: linear second order homogeneous differential equation recursions

  • To: mathgroup at smc.vnet.net
  • Subject: [mg70017] Re: linear second order homogeneous differential equation recursions
  • From: Roger Bagula <rlbagula at sbcglobal.net>
  • Date: Sat, 30 Sep 2006 05:12:50 -0400 (EDT)

 >Daniel Lichtblau wrote:

>Out[14]//InputForm=
>(I^n*2^(-2 + n)*((-I)*Gamma[3/4 - Sqrt[5]/4]*Gamma[3/4 + Sqrt[5]/4] +
>    I*(-1)^n*Gamma[3/4 - Sqrt[5]/4]*Gamma[3/4 + Sqrt[5]/4] +
>    2*Gamma[5/4 - Sqrt[5]/4]*Gamma[5/4 + Sqrt[5]/4] +
>    2*(-1)^n*Gamma[5/4 - Sqrt[5]/4]*Gamma[5/4 + Sqrt[5]/4])*
>   Gamma[3/4 - Sqrt[5]/4 + n/2]*Gamma[3/4 + Sqrt[5]/4 + n/2])/
>  (Gamma[3/4 - Sqrt[5]/4]*Gamma[5/4 - Sqrt[5]/4]*Gamma[3/4 + Sqrt[5]/4]*
>   Gamma[5/4 + Sqrt[5]/4]*Gamma[1 + n])


Daniel Lichtblau,
Thanks very much for your help.
You have gone out of your way to help me
and I appreciate it.
These linear differential equation don't seem to be as easy
as everyone seems to imply they are.
3rd and 4th order solutions aren't as easy as doing the recursions.
By series expansion of the recursion solutions,
one should be able to get a class of solutions to these problems
not available to RSolve.

New last night to test if the form continues to higher orders of a[n-m] :
Clear[a]
f[n_] = n^5 - n^4 - n^3 + n^2 - 1
a[n_] := a[n] = f[n]*a[n - 5]/(n*(n - 1)*(n - 2)*(n - 3)*(n - 4)); a[0] = 
    1; a[1] = 1; a[2] = 1; a[3] = 1; a[4] = 1;
Table[a[n]*n!, {n, 0, 30}]
{1, 1, 2, 6, 24, 2399, 6299, 28222, 169338, 1244136, 213748501, 914608501, 
6392593442, 57693964614, 618160168824, 150820728557099, 895583729570699, 
8513087005239262, 102642351647368962, 1446049101566437896, 
457348626455818450501, 3475580442134239108501, 41792540370077642642282, 
630723188870906581942206, 11015422190043395750131704, 
4280782686277834240870909499, 39647680418065890495986048999, 
576674828014460422853597242102, 10453948615037962591384116861858, 
217888564838736991780849397133576, 100443852591771788566536692522676601}

Roger Bagula


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