bo198214 Wrote:Now - also inspired the Newton formula analogy - I was asking myself whether we cant do a similar thing for the iterative square root \( w \) of \( f \), i.e. \( w\circ w =f \).
Nice idea. I've used f(x) = 2x + x^2 , then
f05 = Ser(x)
f05 = 1/2*(f05 + f (serreverse(f05))) \\ recurse
to arrive at the same result like when using the matrix-root.
:-)
[update] it works also for the dxp(x) = exp(x)-1 function, matching the matrix-method.
For exp(x) Pari/GP cannot give a seriesinversion, so I can't check this [/update]
Gottfried
Gottfried Helms, Kassel

