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***bo198214*** · 07/09/2008, 06:18 PM

andydude Wrote:
Gottfried Wrote:Henryk's formula is (I inserted (x) at f°t)

Hmm ... you mean Woon's formula ...

You can also call it Newton's formula:

$x^t = (x-1+1)^t = \sum_{n=0}^\infty \left(t\\n\right) (x-1)^n = \sum_{n=0}^\infty \left(t\\n\right)\sum_{k=0}^n \left(n\\k\right)(-1)^{n-k} x^k$

Woon just applied this to linear operators $A$ instead of $x$ . This is possible because you dont need the commutativity for those formulas to stay true.

However we cant calculate with iterations as with powers, because the composition is no more right distributive. $(f+g)\circ h=f\circ h+g\circ h$ but generally $f\circ (g+h)\neq f\circ g + f\circ h$ . The binomial formula however relies on full (both side) distributivity. Especially generally
$(f-\text{id})^{\circ n} \neq \sum_{k=0}^n \left(n\\k\right) (-1)^{n-k} f^{\circ k}$

The interesting thing however is that both expansions together are then again valid:
$f^{\circ t} = \lambda^t \sum_{n=0}^\infty \left(t\\n\right) \sum_{k=0}^n \left(n\\k\right) (-1)^{n-k} \frac{f^{\circ k}}{\lambda^k}$ for each $\lambda$ as long as the right side converges.

And I think this is new. Especially that the iteration by this formula is the same as regular iteration (which is sure for elliptic iteration, other cases are still to prove).

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