Re: doubt..

Posted by Asd1995 on Jun 23, 2017; 2:37pm
URL: http://discussion-forum.276.s1.nabble.com/doubt-tp7606172p7606191.html

You can also factorise it as (1-g(x))(1-g(1/x))=1 for all x.

If f(x)=1-g(x), we get f(x) f(1/x)=1, so that f(x)=+- x^n