Consider a polynomial of degree ,
- Prove that if and then there is an degree polynomial such that .
- Prove that is a polynomial of degree for any .
- Prove that if and , then , for a polynomial of degree .
- If has real zeros (i.e., there exist numbers such that ), then for all , and for all .
is a polynomial of degree , and if
for distinct , then
where is then an degree polynomial
Proof. We compute, using the binomial theorem,
Now, we want to group together the coefficients on the like-powers of . So we pull out the terms corresponding to , , etc. This gives us,
In the final line, we rewrote the coefficients as sums to view them a little more concisely. Either way, since and all of the ‘s are constants, we have is some constant for each , say and we have,
Hence, is a polynomial of degree
Proof. From part (b) we know that if is a polynomial of degree , then is also a polynomial of the same degree. Further,
So, by part (a), we have
where is a polynomial of degree . Thus,
But, if is a polynomial of degree , then by part (b) again, so is . Hence,
for a degree polynomial, as requested
Proof. The proof is by induction. First, we consider the case in which . Then . Since the assumption is that there are distinct real such that , we know there exist such that
Hence, the statement is true for . Assume that it is true for some . Then, let be a polynomial of degree with distinct real zeros, . Then, since , using part (c), we have,
where is a polynomial of degree . Further, we know there are distinct values such that . (Since for and for with since all of the are distinct.) Thus, by the induction hypothesis, every coefficient of is 0 and for all . Thus,
But, since all of the coefficients of are zero, we have for . Hence,
Thus, all of the coefficients of are zero and for all . Hence, the statement is true for the case , and therefore, for all
where for (we have by assumption).
Then, there are distinct real for which . (Since there are distinct real values for which , and so at each of these values .) Thus, by part (d), for and for all . But then,
for all . Further, since for , and by assumption for , we have for . But then,
means is a polynomial of degree as well