flinn@seismo.UUCP (E. A. Flinn) (12/02/83)
The following problem appeared in Technology Review recently: Solve the infinite-order ordinary differential equation f = f' + 2f" + 3f'" + 4f"" + ...
mkr@CS-Arthur (Mahesh K Rathi) (12/04/83)
The solution to the infinite order differential equation f = f'+2f''+3f'''+ ...... is f(x) = exp(rx) where r = (3-sqrt(5))/2, which is easily obtained by putting f=exp(rx) in the diff. eqn. How did I guess that the solution should be of the type f=exp(rx) ? Well, by noting that the soln. to f=f' is exp(x), the soln. to f=f'+2f'' is exp(x/2) etc. etc.
wolit@rabbit.UUCP (12/05/83)
If f(x) = 0, then f = f' + 2f" + 3f'" + 4f"" + ... = 0 Jan Wolitzky, AT&T Bell Labs, Murray Hill, NJ
apdoo@alice.UUCP (12/05/83)
There is, of course, a symbolic way of solving this equation. Writing D for the differential operator, we have inf n f=sum nD f n=1 so Df f=--------- (1-D)**2 Solving, (1-D)**2 f =Df f-3Df+D**2 f =0 f=c exp(rx) + k exp(sx) where c and k are arbitrary constants and r and s are the roots of 1-3x+x**2=0 or (3+-sqrt(5))/2