- 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 :)