Thread Rating:
  • 1 Vote(s) - 5 Average
  • 1
  • 2
  • 3
  • 4
  • 5
How about ... F(x)^[3] + 2 F(x)^[2] = exp(x) ?
#1
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
Reply
#2
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
Reply
#3
Maybe start the iteration with exp^[1/3](x).

Regards

Tommy1729
Reply
#4
(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
Reply




Users browsing this thread: 1 Guest(s)