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 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
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 aredistinct 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
; 
.\footnote{It is not strictly necessary to split the domain
is zero whenever
and
} This allows decoupling of the summations. Since
,
when
, hence
when
like this since under the generalised definition of the binomial
coefficient,
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
\[
p(x) = f(x+a) = \sum_{k=0}^{n}c_k(x+a)^k =\
\sum_{k=0}^{n}\sum_{j=0}^kc_k\binom{k}{j}x^ja^{k-j}
\]
Let \(d_{j,k}=c_k\binom{k}{j}a^{k-j}\) 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
\[
\sum_{k=0}^{n}\sum_{j=0}^kc_k\binom{k}{j}x^ja^{k-j}=\
\sum_{k=0}^{n}\sum_{j=0}^nd_{j,k}x^j
\]
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 n-1.
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!