There is a theorem that states that if f(x) and g(x) are rational functions (and g(x) is not a constant) , then can be expressed in terms of a finite number of elementary functions iff there exists a rational function R(x) such that .
So let's take
So I have to show that there does not exist a R(X) such that .
It's a linear first order differential equation with integrating factor
How do I show that R(x) is not rational? Do I have to show that is nonelementary?
thanks mr f.
suppose otherwise. then there exists a rational function where are some polynomials in and obviously we may assume that
have no common zeros. we have thus: which gives us:
now consider two cases:
case 1: : in this case would be a polynomial satisfying this is impossible because
case 2: let where and put this in to get: which is impossible
because putting will give us and so would be a common zero of contradiction! Q.E.D.
Is it possible to write and/or so that they are in the proper form to use the theorem?
My intuition tells me to use the complex form of for . But then I have the difference of two functions of the proper form.
I have no idea what to do with . Unless my understaning of the definition of a rational function is incorrect (namely that a rational function is function that can be written as the quotient of two polynomials) I can't writex as