[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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