Let
- Prove that
where denotes the th derivative of .
- Do part (a) in the case that is a cubic polynomial.
- Find a similar formula and prove it in the case that is a polynomial of degree .
For all of these we recall from a previous exercise (Section 5.11, Exercise #4) that by Leibniz’s formula if then the th derivative is given by
So, in the case at hand we have and so
(Since the th derivative of is still for all and .)
- Proof. From the formula above we have
But, since is a quadratic polynomial we have
Hence, we have
- If is a cubic polynomial we may write,
Claim: If then
Proof. We follow the exact same procedure as part (a) except now we have the derivatives of given by
Therefore, we now have
- Claim: Let be a polynomial of degree ,
Let . Then,
Proof. Using Leibniz’s formula again, we have
But for the degree polynomial , we know if and for all . Hence, we have