341or let g(x) = (x-1)p(x)
Then g(x+1) - g(x)= 1 which from the theory of finite differences implies that g(x) is of degree 1.
which means P(x) is a constant = c
c(x-1)+1 = cx implies c =1
\hspace{-16}$Find all Polynomials $p(x)$ such that $(x-1).p(x)+1=x.p(x+1)$
-
UP 0 DOWN 0 0 3
3 Answers
21Let g(x)= p(x)-1, the given assertion becomes x\cdot g(x+1)= (x-1)\cdot g(x), Note that g(2)= g(0)= 0, and if for an integer t>2, g(t)= 0 \implies g(t+1)= 0, So g(x)= 0 \forall x \in \mathbb{R}, i.e p(x)= 1 is the only polynomial satisfying the given relation.
341