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 How about ... F(x)^[3] + 2 F(x)^[2] = exp(x) ? tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 03/06/2016, 11:44 PM 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 tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 03/07/2016, 11:07 AM 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 tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 03/07/2016, 09:07 PM Maybe start the iteration with exp^[1/3](x). Regards Tommy1729 tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 11/10/2021, 12:26 AM (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 « Next Oldest | Next Newest »

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