- Find a polynomial satisfying
Prove that there is only one such polynomial.
- Given a polynomial , prove there is exactly one polynomial such that
- Proof. (Finding the polynomial will prove that it is unique since we will not have any choices to make while deriving the polynomial .) First, we write
Thus, we have
Setting this equal to we have
But, this implies and since is the only term on the left (so if , then we couldn’t have the largest power of on the right). Therefore, is a degree polynomial and with , so we have
Hence we have
Thus we have the equations
These uniquely determine and ,
Hence, there is a unique satisfying this equation,
- Proof. Let be a given polynomial and suppose there exist two polynomials and such that
This implies
Now, if then it is of degree for some . We know its derivative has degree (Apostol, Page 166). But then, this would imply
has degree (since the coefficient of in is zero since it is degree , and the coefficient of is nonzero since it has degree ). But we know this difference is 0, which means it cannot have degree for any . Thus, we must have or
Not to be a pedant, but for part b you would need to show that at least one solution exists before you can proceed to the step of showing that any two solutions must be equal. Though this could be easily shown like in part a, so if you were like me, you were just too lazy to have to write that all out in LaTeX :)