Assume is a polynomial with
. Define
Compute the values of .
Assume is a non-negative integer (otherwise
is undefined at
). Then, we make the following claim:
Claim: The polynomial has derivatives at 0 given by the following
Proof. Since is a polynomial we may write,
Furthermore, since is given we know
. Now, multiplying by
we have
Next, we will use induction to prove that the th derivative of
for
is given by
for constants . Since the derivative
is given by
for constants , we see that the formula holds for
. Assume then that it holds for some
,
Then, taking the derivative of this we have,
Hence, the formula holds for all . But then, if
we have
If then
for all
; hence,