Consider the function
and .
- 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
Therefore,
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