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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.

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Tommy1729

Posts: 1,493

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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 ?

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Tommy1729

Posts: 1,493

Threads: 356

Joined: Feb 2009

Maybe start the iteration with exp^[1/3](x).

Regards

Tommy1729

Posts: 1,493

Threads: 356

Joined: Feb 2009

(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