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 .
 If
is a polynomial of degree , and if
for distinct , then

Proof. First,
Thus,
where is then an degree polynomial

Proof. We compute, using the binomial theorem,
Now, we want to group together the coefficients on the likepowers 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
Thus,
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

Proof. Let
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,and
for all . Further, since for , and by assumption for , we have for . But then,
means is a polynomial of degree as well
\documentclass{article}
\usepackage{amsmath}
\begin{document}
Alternative proof for part b:
We have
Let when ;
when .\footnote{It is not strictly necessary to split the domain
like this since under the generalised definition of the binomial
coefficient, is zero whenever and
} This allows decoupling of the summations. Since , when , hence
Now the summations are decoupled, they can be reversed. This can be
proven by induction or simply by considering the summation of all
numbers in a square table.
Finally, let . This gives us
which is a polynomial of degree n as per the definition.
\end{document}
\documentclass{article}
\usepackage{amsmath}
\begin{document}
Alternative proof for part b:
We have
\[
p(x) = f(x+a) = \sum_{k=0}^{n}c_k(x+a)^k =\
\sum_{k=0}^{n}\sum_{j=0}^kc_k\binom{k}{j}x^ja^{kj}
\]
Let \(d_{j,k}=c_k\binom{k}{j}a^{kj}\) when \(j\leq k\); \(d_{j,k}=0\)
when \(j>k\).\footnote{It is not strictly necessary to split the domain
like this since under the generalised definition of the binomial
coefficient, \(\binom{k}{j}\) is zero whenever \(j>k\) and
\(j,k\in\mathbb{Z}\)} This allows decoupling of the summations. Since \(k\leq
n\), \(d_{j,k}=0\) when \(j>n\), hence
\[
\sum_{k=0}^{n}\sum_{j=0}^kc_k\binom{k}{j}x^ja^{kj}=\
\sum_{k=0}^{n}\sum_{j=0}^nd_{j,k}x^j
\]
Now the summations are decoupled, they can be reversed. This can be
proven by induction or simply by considering the summation of all
numbers in a square table.
\[
\sum_{k=0}^{n}\sum_{j=0}^nd_{j,k}x^j=\
\sum_{j=0}^{n}\sum_{k=0}^nd_{j,k}x^j=\
\sum_{j=0}^{n}x^j\sum_{k=0}^nd_{j,k}
\]
Finally, let $e_j=\sum_{k=0}^nd_{j,k}$. This gives us
\[
p(x)=\sum_{j=0}^{n}e_jx^j
\]
which is a polynomial of degree n as per the definition.
\end{document}
Hi, as always great work!
Just a couple of points:
In d), just before the last step, the sign for the last term is minus and the indices of the term before are switched it is [c(0)c(1)*a(k+2)]x.
An interesting observation in b) is that by collecting all the terms with x^0 (i.e. the constant coefficient of p(x)) we are writing the original function f(x) with x=a i.e. f(a).
Hence, if ‘a’ is a root of f(x) then by default the constant coefficient of p(x) will be zero. It immediately follows that p(x) can be rewritten as p(x) = x(gx), where g(x) is a polynomial of degree n1.
Hey, could you please explain what happens between lines 4 and 5 of question 9 (b)? It’s not leaping out at me, sorry :s
Hi, yeah, I wrote that in sort of a confusing way, and there’s at least one mistake (an that should be ). So, starting with line 4 we have
(Where we go to that by grouping together the like powers of from the previous line.) We could actually just stop there and say each of those coefficients is some constant (since and all of the are constants). But, for some reason (which eludes me at the moment) I wanted to write them down as a sum. So, to do that I really just observed what was in those coefficients and wrote them down in summation notation. So, what’s written there is
If we expand out the terms we have
Which was what we had on the line before… so all that has actually happened is that I rewrote those coefficients in summation notation. As to how I arrived at that summation formula… I probably just stared at it for a while until I figured out what the sums needed to be. I don’t think there was any smart way. Maybe it would be better to just leave that part off since we already knew those were constants, which was what we wanted to show.
Does that help?
Thank you for your solution! The last step was just frustrating for me!
I updated the post for part (b) a bit (and fixed the typo). Maybe it is a bit more clear now? I’m not sure if it’s all that much better.
Yup, that step where you expanded the terms was the step I needed! Thanks a lot, I’m working through Apostol right now, and your blog has been incredibly useful!