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+--- Thread: How about ... F(x)^[3] + 2 F(x)^[2] = exp(x) ? (/showthread.php?tid=1069)
How about ... F(x)^[3] + 2 F(x)^[2] = exp(x) ? - tommy1729 - 03/06/2016
How about ... F(x)^[3] + 2 F(x)^[2] = exp(x) ?
I have no idea how to do this.
This bugs me. Its a simple equation.
Regards
Tommy1729
RE: How about ... F(x)^[3] + 2 F(x)^[2] = exp(x) ? - tommy1729 - 03/07/2016
It seems logical to consider F_{n+1}(x) = ( 1/2 ( exp(x) - f_{n}^[3](x) ) )^[1/2].
The limit of the iteration is then the solution.
But how about convergeance and analyticity ?
Regards
Tommy1729
RE: How about ... F(x)^[3] + 2 F(x)^[2] = exp(x) ? - tommy1729 - 03/07/2016
Maybe start the iteration with exp^[1/3](x).
Regards
Tommy1729
RE: How about ... F(x)^[3] + 2 F(x)^[2] = exp(x) ? - tommy1729 - 11/10/2021
(03/07/2016, 11:07 AM)tommy1729 Wrote: It seems logical to consider F_{n+1}(x) = ( 1/2 ( exp(x) - f_{n}^[3](x) ) )^[1/2].
The limit of the iteration is then the solution.
But how about convergeance and analyticity ?
Regards
Tommy1729
I see no reason why this would fail , assuming a good starting function and starting point x.
This should be investigated imo.
regards
tommy1729
Tom Marcel Raes