[seqfan] Re: Fibonacci-sums

Gottfried Helms Annette.Warlich at t-online.de
Thu Aug 19 16:17:34 CEST 2010


The matrix-trick allows even more simple answers to the
questions of my first post.

We had, that

  as = fib(0) - fib(1) + fib(2) - fib(3) + ...

could be computed by the closed form of the alternating geometric
series of the matrix m0

  AS = ( I + m0) ^-1  = [ 2  1 ] ^-1  = [ 1  -1 ]
                        [ 1  1 ]        [-1   2 ]

 and
   as = AS[2,1] = -1

-----------------------------------------------------------------

For the following it is better to correct for the initial values.
So from now I discuss

  as = fib(1) - fib(2) + fib(3) - fib(4) + ...

which requires

  AS = m0 * ( I + m0)^-1

**  This inverts the sign of the previously computed value, meaning
**  we do not get as = -1 but now we get as=1


In my previous posting I asked for generalization, this means now:

           n             n             n             n
  as(n) = 1  * fib(1) - 2  * fib(2) + 3  * fib(3) - 4 * fib(4) + ...

Now because the n'th powers of consecutive integers as coefficients
of the formal powerseries occur by composition of the derivatives
of x/(1 + x), we get the following closed-form-expressions for
the matrix-geometric-series and their derivatives
(where I is the identity matrix)

   AS(0) = m0*( I + m0) ^-1                                         => as(0)=   1
   AS(1) = m0*( I + m0) ^-2   * ( 1*I )                             => as(1)=   2
   AS(2) = m0*( I + m0) ^-3   * ( 1*I - 1*m0 )                      => as(2)=   8
   AS(3) = m0*( I + m0) ^-4   * ( 1*I - 4*m0 + 1*m0^2 )             => as(3)=  50
   AS(4) = m0*( I + m0) ^-5   * ( 1*I -11*m0 +11*m0^2 - 1*m0^3 )    => as(4)= 416

 and so on

where the coefficients in the rhs-triangle are taken of the triangle
of Eulerian numbers. The values as(m) can then be taken from the value in AS(m)[2,1]



Very nice... The more I'm now interested, whether we can extend this to
the non-alternating sums

Gottfried Helms




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