That is right, but you wrote in a previous post: " If Q in Rn[x] then there is a P in Rn-1[x] such that f(Q)=P " , which is not the same thing.
Fernando Revilla
[QUOTE=FernandoRevilla;620861]Right if you change by .
One final question: What do you mean change P by Q? If P is in Rn-1[X] then Q is in Rn[X] such that fn(Q)=P. Why chanige P by Q in my reasoning?
Thanks for all the information, guidance until now Dr. Revilla! I appreciate everything!
That is right, but you wrote in a previous post: " If Q in Rn[x] then there is a P in Rn-1[x] such that f(Q)=P " , which is not the same thing.
Fernando Revilla