Fundamental properties of polynomials

Consider a polynomial of degree $n$ ,

$f(x) = \sum_{k=0}^n c_k x^k = c_0 + c_1 x + \cdots + c_n x^n.$

Prove that if $f(0) = 0$ and $n \geq 1$ then there is an $n-1$ degree polynomial $g(x)$ such that $f(x) = x g(x)$ .
Prove that $p(x) = f(x+a)$ is a polynomial of degree $n$ for any $a \in \mathbb{R}$ .
Prove that if $n \geq 1$ and $f(a) = 0$ , then $f(x) = (x-a)h(x)$ , for a polynomial $h(x)$ of degree $n-1$ .
If $f(x)$ has $n+1$ real zeros (i.e., there exist $n+1$ numbers $x \in \mathbb{R}$ such that $f(x) = 0$ ), then $c_k =0$ for all $k = 0, \ldots, n$ , and $f(x) = 0$ for all $x \in \mathbb{R}$ .
If
$g(x) = \sum_{k=0}^m b_k x^k$

is a polynomial of degree $m \geq n$ , and if

$g(x) = f(x)$

for $m+1$ distinct $x \in \mathbb{R}$ , then

$m=n, \qquad b_k = c_k \quad \text{for each } k, \qquad g(x) = f(x) \quad \text{for all } x \in \mathbb{R}.$

Proof. First,

$f(0) = 0 \quad \implies \quad c_0 + c_1 \cdot 0 + \cdots + c_{n-1} \cdot 0^{n-1} + c_n 0^n = 0 \quad \implies \quad c_0 = 0.$

Thus,

$\begin{align*} f(x) &= c_1x + c_2 x^2 + \cdots + c_n x^n \\ &= x (c_1 + c_2 x + \cdots + c_n x^{n-1}) \\ &= x g(x) \end{align*}$

where $g(x) = \sum_{k=0}^{n-1} c_{k+1} x^k$ is then an $n-1$ degree polynomial $. \qquad \blacksquare$
Proof. We compute, using the binomial theorem,

$\begin{align*} p(x) = f(x+a) &= \sum_{k=0}^n (x+a)^k c_k \\ &= c_0 + (x+a)c_1 + (x+a)^2 c_2 + \cdots + (x+a)^n c_n \\ &= c_0 + c_1 \left(\sum_{j=0}^1 \binom{1}{j} a^j x^{1-j}\right) + c_2 \left( \sum_{j=0}^2 \binom{2}{j} a^j x^{2-j} \right) + \cdots + c_n \left( \sum_{j=0}^n \binom{n}{j} a^j x^{n-j} \right) \end{align*}$

Now, we want to group together the coefficients on the like-powers of $x$ . So we pull out the terms corresponding to $x^0$ , $x^1$ , etc. This gives us,

$\begin{align*} &= (c_0 + ac_1 + a^2 c_2 + \cdots + a^n c_n) + x(c_1 + 2ac_2 + \cdots + na^{n-1}c_n) \\ &\qquad \qquad + x^2\left(c_2 + 3ac_3 + \cdots + \binom{n}{n-2}a^{n-2} c_n \right) + \cdots + x^n c_n \\ &= \sum_{k=0}^n \left( x^k \left( \sum_{j=k}^n \binom{j}{j-k} c_j a^{j-k} \right) \right). \end{align*}$

In the final line, we rewrote the coefficients as sums to view them a little more concisely. Either way, since $a$ and all of the $c_i$ ‘s are constants, we have $\sum_{j=k}^n \binom{j}{j-k} c_j a^{j-k}$ is some constant for each $k$ , say $d_k$ and we have,

$p(x) = \sum_{k=0}^n d_k x^k.$

Hence, $p(x)$ is a polynomial of degree $n. \qquad \blacksquare$
Proof. From part (b) we know that if $f(x)$ is a polynomial of degree $n$ , then $p(x) = f(x+a)$ is also a polynomial of the same degree. Further,

$f(a) = 0 \quad \implies \quad p(0) = f(a) = 0.$

So, by part (a), we have

$p(x) = x \cdot g(x)$

where $g(x)$ is a polynomial of degree $n-1$ . Thus,

$p(x-a) = f(x) = (x-a)\cdot g(x-a).$

But, if $g(x)$ is a polynomial of degree $n-1$ , then by part (b) again, so is $h(x) = g(x+(-a))= g(x-a)$ . Hence,

$f(x) = (x-a) \cdot h(x)$

for $h$ a degree $n-1$ polynomial, as requested $. \qquad \blacksquare$
Proof. The proof is by induction. First, we consider the case in which $n = 1$ . Then $f(x) = c_0 + c_1 x$ . Since the assumption is that there are $n+1$ distinct real $x$ such that $f(x) = 0$ , we know there exist $a_1, a_2 \in \mathbb{R}$ such that

$f(a_1) = f(a_2) = 0, \qquad a_1 \neq a_2.$

Thus,

$\begin{align*} c_0 + c_1 a_1 = c_0 + c_1 a_2 = 0 && \implies && c_1 a_1 - c_1 a_2 &= 0 \\ && \implies && c_1 (a_1 - a_2) &= 0 \\ && \implies && c_1 &= 0 & (\text{since } a_1 \neq a_2)\\ && \implies && c_0 &= 0 & (\text{since } c_0 + c_1 a_1 = 0). \end{align*}$

Hence, the statement is true for $n=1$ . Assume that it is true for some $n = k \in \mathbb{Z}_{>0}$ . Then, let $f(x)$ be a polynomial of degree $k+1$ with $k+2$ distinct real zeros, $a_1, \ldots, a_{k+2}$ . Then, since $f(a_{k+2}) = 0$ , using part (c), we have,

$f(x) = (x - a_{k+2})h(x)$

where $h(x)$ is a polynomial of degree $k$ . Further, we know there are $k+1$ distinct values $a_1, \ldots, a_{k+1}$ such that $h(a_i) = 0$ . (Since $f(a_i) = 0$ for $1 \leq i \leq k+2$ and $(x - a_{k+2}) \neq 0$ for $x = a_i$ with $1 \leq i \leq k+1$ since all of the $a_i$ are distinct.) Thus, by the induction hypothesis, every coefficient of $h$ is 0 and $h(x) = 0$ for all $x \in \mathbb{R}$ . Thus,

$\begin{align*} f(x) = (x - a_{k+2}) h(x) &= (x-a_{k+2}) \cdot \sum_{j=0}^k c_j x^j \\ &= \sum_{j=0}^k (x-a_{k+2})c_j x^j \\ &= c_k x^{k+1} + (c_{k-1} - a_{k+2} c_k)x^k + \cdots + (c_1 - a_{k+2} c_0) x + a_{k+2} c_0. \end{align*}$

But, since all of the coefficients of $h(x)$ are zero, we have $c_i = 0$ for $i = 1, \ldots k$ . Hence,

$f(x) = 0 \cdot x^{k+1} + 0 \cdot x^k + \cdots + 0 \cdot x + 0 = 0 \qquad \text{for all } x \in \mathbb{R}.$

Thus, all of the coefficients of $f$ are zero and $f(x) = 0$ for all $x \in \mathbb{R}$ . Hence, the statement is true for the case $k+1$ , and therefore, for all $n \in \mathbb{Z}_{>0}. \qquad \blacksquare$
Proof. Let

$p(x) = g(x) - f(x) = \sum_{k=0}^m b_k x^k - \sum_{k=0}^n c_k x^k = \sum_{k=0}^m (b_k - c_k)x^k$

where $c_k = 0$ for $n < k \leq m$ (we have $m \geq n$ by assumption).
Then, there are $m+1$ distinct real $x$ for which $p(x) = 0$ . (Since there are $m+1$ distinct real values for which $g(x) = f(x)$ , and so at each of these values $p(x) = g(x) - f(x) = 0$ .) Thus, by part (d), $b_k - c_k = 0$ for $k = 0, \ldots, m$ and $p(x) = 0$ for all $x \in \mathbb{R}$ . But then,

$b_k - c_k = 0 \quad \implies \quad b_k = c_k \qquad \text{for }k =0, \ldots, m$

and

$p(x) = 0 \quad \implies \quad g(x) - f(x) = 0 \quad \implies \quad f(x) = g(x),$

for all $x \in \mathbb{R}$ . Further, since $b_k - c_k = 0$ for $k = 0, \ldots, m$ , and by assumption $c_k = 0$ for $k = n+1, \ldots, m$ , we have $b_k = 0$ for $k = n+1, \ldots, m$ . But then,

$g(x) = \sum_{k=0}^n b_k x^k + \sum_{k=n+1}^m 0 \cdot x^k = \sum_{k=0}^n b_k x^k$

means $g(x)$ is a polynomial of degree $n$ as well $. \qquad \blacksquare$

April 19, 2016 at 9:44 am

RoRi says:

Hi, yeah, I wrote that in sort of a confusing way, and there’s at least one mistake (an $a_n$ that should be $a^n$ ). So, starting with line 4 we have

$\begin{align*} (c_0 + ac_1 + a^2c_2 + \cdots + a^n c_n) + x (c_1 + 2ac_2 + \cdots + na^{n-1} c_n) \\ + x^2 \left( c_2 + 3ac_3 + \cdots + \binom{n}{n-2} a^{n-2} c_n \right) + \cdots + x^n c_n. \end{align*}$

(Where we go to that by grouping together the like powers of $x$ from the previous line.) We could actually just stop there and say each of those coefficients is some constant $d_k$ (since $a$ and all of the $c_0, \ldots, c_n$ 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

$\sum_{k=0}^n \left( x^k \left( \sum_{j=k}^n \binom{j}{j-k} c_j a^{j-k} \right) \right).$

If we expand out the terms we have

$\begin{align*} \sum_{k=0}^n \left( x^k \left( \sum_{j=k}^n \binom{j}{j-k} c_j a^{j-k} \right) \right) &= x^0 \left( \sum_{j=0}^n \binom{j}{j} c_j a^{j} \right) + x^1 \left( \sum_{j=1}^n \binom{j}{j-1} c_j a^{j-1} \right) + \cdots + x^n \left( \sum_{j=n}^n \binom{j}{j-n} c_j a^{j-n} \right) \\[11pt] &= x^0 (c_0 + c_1 a + \cdots + c_n a^n) + x (c_1 + 2c_2 a + \cdots + nc_n a^{n-1}) + \cdots + x^n (c_n). \end{align*}$

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?

8 comments

January 14, 2018 at 9:22 am

Jon Hurst says:

\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}

January 14, 2018 at 9:21 am

Jon Hurst says:

\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}

June 27, 2017 at 1:40 pm

Claudio Rebelo says:

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.

April 19, 2016 at 12:48 am

Darcy says:

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

- April 19, 2016 at 9:44 am
  
  RoRi says:
  
  Hi, yeah, I wrote that in sort of a confusing way, and there’s at least one mistake (an $a_n$ that should be $a^n$ ). So, starting with line 4 we have
  
  $\begin{align*} (c_0 + ac_1 + a^2c_2 + \cdots + a^n c_n) + x (c_1 + 2ac_2 + \cdots + na^{n-1} c_n) \\ + x^2 \left( c_2 + 3ac_3 + \cdots + \binom{n}{n-2} a^{n-2} c_n \right) + \cdots + x^n c_n. \end{align*}$
  
  (Where we go to that by grouping together the like powers of $x$ from the previous line.) We could actually just stop there and say each of those coefficients is some constant $d_k$ (since $a$ and all of the $c_0, \ldots, c_n$ 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
  
  $\sum_{k=0}^n \left( x^k \left( \sum_{j=k}^n \binom{j}{j-k} c_j a^{j-k} \right) \right).$
  
  If we expand out the terms we have
  
  $\begin{align*} \sum_{k=0}^n \left( x^k \left( \sum_{j=k}^n \binom{j}{j-k} c_j a^{j-k} \right) \right) &= x^0 \left( \sum_{j=0}^n \binom{j}{j} c_j a^{j} \right) + x^1 \left( \sum_{j=1}^n \binom{j}{j-1} c_j a^{j-1} \right) + \cdots + x^n \left( \sum_{j=n}^n \binom{j}{j-n} c_j a^{j-n} \right) \\[11pt] &= x^0 (c_0 + c_1 a + \cdots + c_n a^n) + x (c_1 + 2c_2 a + \cdots + nc_n a^{n-1}) + \cdots + x^n (c_n). \end{align*}$
  
  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?
  
  - January 17, 2017 at 2:06 am
    
    Anonymous says:
    
    Thank you for your solution! The last step was just frustrating for me!
- April 19, 2016 at 9:53 am
  
  RoRi says:
  
  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.
  
  - April 20, 2016 at 11:14 pm
    
    Darcy says:
    
    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!

Stumbling Robot

Fundamental properties of polynomials

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