Consider the function
- Prove that for every positive number we have
- Prove that if then
where is a polynomial in .
- Prove that
- Proof. (A specific case of this general theorem is actually the first problem of this section, here. Maybe it’s worth taking a look since this proof is just generalizing that particular case.) We make the substitution , so that as and we have
by Theorem 7.11 (page 301 of Apostol) since implies as well
- Proof. The proof is by induction on . In the case we have
So, indeed the formula is valid in the case . Assume then that the formula holds for some positive integer . We want to show this implies the formula holds for the case .
But then the leading term
is still a polynomial in since the derivative of a polynomial in is still a polynomial in , and so is the sum of two polynomials in . Therefore, we have that the formula holds for the case ; hence, it holds for all positive integers
- Proof. The proof is by induction on . If then we use the limit definition of the derivative to compute the derivative at 0,
So, indeed and the statement is true for the case . Assume then that for some positive integer . Then, we use the limit definition of the derivative again to compute the derivative ,
This follow since is still a polynomial in , and by the definition of for . But then, by part (a) we know
Thus, the formula holds for the case , and hence, for all positive integers
The last step is not clear, and it is safer to substitute x=1/t and apply t7.11.
Using a sum for p(1/x) with part b should make things clear for you