#### prerequisites: A basic knowledge of Haskell or similar.

“What makes the desert beautiful,”
said the little prince,
“is that somewhere it hides a well…”
Antoine de Saint-Expuéry, The Little Prince

### 1. Introduction

The goal of this blog post is to introduce Horner’s method for polynomial evaluation and polynomial division, and subsequently prove an equivalence relation between these two types of application.

The blog post is structured as follows. In Section 2, we argue for the application of Horner’s method for polynomial evaluation, and subsequently derive its definition. Having covered polynomial evaluation, we then argue for the application of Horner’s method for polynomial division and derive its definition in Section 3. Lastly, we state and prove an equivalence relation between the definition of polynomial evaluation and the definition of polynomial division using Horner’s method in Section 4. The blog post is concluded in Section 5.

### 2. Polynomial evaluation using Horner’s method

In order to understand the advantages of using Horner’s method for evaluating a polynomial, we first examine how this is usually done. If we let $$p(x) = 7x^4 + 2x^3 + 5x^2 + 4x + 6$$ and $$x = 3$$, then we would evaluate $$p(3)$$ one term at a time and sum all the intermediate results. However, by doing so we are unfortunately performing redundant operations when evaluating the exponents, as can be seen when we unfold the evaluation of the exponents,

\begin{align*} p(3) &= 7 \cdot (3^4) + 2 \cdot (3^3) + 5 \cdot (3^2) + 4 \cdot (3) + 6 \\ &= 7 \cdot (3 \cdot 3 \cdot 3 \cdot 3) + 2 \cdot (3 \cdot 3 \cdot 3) + 5 \cdot (3 \cdot 3) + 4 \cdot (3) + 6. \end{align*}

Here, the evaluation of the largest exponent, $$3^4 = (3 \cdot 3 \cdot 3 \cdot 3)$$, also calculates all exponents of a lesser degree, i.e., $$3^3 = (3 \cdot 3 \cdot 3)$$ and $$3^2 = (3 \cdot 3)$$, as its intermediate results. Luckily, we can transform the formula of the polynomial $$p$$ in such a way that the operations calculating the exponents are shared across the terms. In fact, since the number of multiplications by $$3$$ decreases by $$1$$ for each term, we can nest the multiplications across the terms like so,

\begin{align} \tag{1}\label{eq:polynomial-evaluation-horner-example-formula} p(3) &= 7 \cdot (3 \cdot 3 \cdot 3 \cdot 3) + 2 \cdot (3 \cdot 3 \cdot 3) + 5 \cdot (3 \cdot 3) + 4 \cdot (3) + 6 \\ p(3) &= (7 \cdot (3 \cdot 3 \cdot 3) + 2 \cdot (3 \cdot 3) + 5 \cdot (3) + 4) \cdot 3 + 6 \\ p(3) &= ((7 \cdot (3 \cdot 3) + 2 \cdot (3) + 5) \cdot 3 + 4) \cdot 3 + 6 \\ p(3) &= (((7 \cdot 3 + 2) \cdot 3 + 5) \cdot 3 + 4) \cdot 3 + 6, \end{align}

thus removing any redundant multiplications used for evaluating the exponents. Formula \ref{eq:polynomial-evaluation-horner-example-formula} now exhibits a simple inductive structure which adds one multiplication and one addition for each term in the polynomial $$p$$. As a result, we can now evaluate the final formula of $$p$$,

$$$\tag{2}\label{eq:polynomial-evaluation-horner-example-inductive} p(3) = (((7 \cdot 3 + 2) \cdot 3 + 5) \cdot 3 + 4) \cdot 3 + 6,$$,$

by repeatedly performing a multiplication and an addition, starting from the innermost set of parentheses,

\begin{align} \tag{3}\label{eq:polynomial-evaluation-horner-example-calculation} p(3) &= (((7 \cdot 3 + 2) \cdot 3 + 5) \cdot 3 + 4) \cdot 3 + 6 \\ &= ((23 \cdot 3 + 5) \cdot 3 + 4) \cdot 3 + 6 \\ &= (74 \cdot 3 + 4) \cdot 3 + 6 \\ &= 226 \cdot 3 + 6 \\ &= 684. \end{align}

If we compare the number of operations performed in the first and last equation of Formula \ref{eq:polynomial-evaluation-horner-example-formula}, we count a total of $$14$$ in the former and $$8$$ in the latter. The difference of $$6$$ operations corresponds exactly to the number of multiplications required to evaluate the exponents in the first equation. Thus, our transformation of the polynomial formula into its inductive form, has removed the computational overhead of evaluating each of the exponents in sequence. Lastly, it has even been proved that the number of additions and multiplications used in this procedure, are indeed the smallest number possible for evaluating a polynomial.1

Upon closer examination of the intermediate results of Formula \ref{eq:polynomial-evaluation-horner-example-calculation}, we can make out a recursive substitution scheme happening under the hood,

\begin{align} \tag{4}\label{eq:polynomial-evaluation-horner-example-substitution-numbers} 7 &= 7\\ 23 &= 7 \cdot 3 + 2\\ 74 &= 23 \cdot 3 + 5\\ 226 &= 74 \cdot 3 + 4\\ 684 &= 226 \cdot 3 + 6. \end{align}

where each intermediate result is the result of multiplying the previous result by $$3$$ and adding the next coefficient. If we assign the intermediate values on the left-hand side, $$(7, 23, 74, 226, 684)$$, to the variable $$b_i$$, assign the value $$3$$ to the variable $$k$$, and lastly assign the values corresponding to the coefficients of $$p$$, $$(7, 2, 5, 4, 6)$$, to the variable $$a_i$$, we can restate Formula \ref{eq:polynomial-evaluation-horner-example-substitution-numbers} like so,

\begin{align} \tag{5}\label{eq:polynomial-evaluation-horner-example-substitution-variables} b_4 &= a_4\\ b_3 &= b_4 \cdot k + a_3\\ b_2 &= b_3 \cdot k + a_2\\ b_1 &= b_2 \cdot k + a_1\\ b_0 &= b_1 \cdot k + a_0. \end{align}

Formula \ref{eq:polynomial-evaluation-horner-example-substitution-variables} now reflects a recursively structured, and easily generalizable, substitution procedure where $$b_4 = a_4$$ is the base case, and the inductive case is defined in terms of the next coefficient in the polynomial and the preceding intermediate result, $$b_3 = b_4 \cdot k + a_3$$. The procedure terminates when it reaches the last term of the polynomial $$p$$, where $$b_0 = b_1 \cdot k + a_0$$ is the result of evaluating $$p(k)$$.

We call the above procedure Horner’s method2 for polynomial evaluation, and formalize it in Haskell by first representing a polynomial as a list of integers,

for which we define the procedure,

which takes a polynomial, cs, corresponding to $$a_i$$, an integer, x, corresponding to $$k$$, and an accumulator, a, corresponding to the intermediate result $$b_i$$. As described above, it returns the final value of the accumulator (result), a, in the base case, and multiplies a by x for each recursive call and adds the coefficient c. Lastly, we define a wrapper procedure,

which initializes the accumulator to 0. As a result, we now can evaluate the example polynomial of Formula \ref{eq:polynomial-evaluation-horner-example-inductive},

$\begin{equation*} 7x^4 + 2x^3 + 5x^2 + 4x + 6, \end{equation*}$

for $$x = 3$$, by passing the coefficients of $$p$$ as the list [7, 2, 5, 4, 6], along with the value of $$x$$, 3, to hornersPolyEval like so, hornersPolyEval [7, 2, 5, 4, 6] 3, giving the expected result, 684.

Having formalized Horner’s method for polynomial evaluation, as the procedures hornersPolyEvalAcc and hornersPolyEval, we now define Horner’s method for polynomial division.

### 3. Polynomial division using Horner’s method

Now that we have used Horner’s method as an efficient procedure for evaluating a polynomial, using a recursive substitution scheme, we move on to examine its use for polynomial division.

According to the definition of polynomial division, when dividing two polynomials, $$p$$ and $$d$$, $$\frac{p(x)}{d(x)}$$, where $$d \not= 0$$, the result is a quotient, $$q$$, and a remainder, $$r$$, satisfying the relation,

$$$\tag{6}\label{eq:poly-div-relation} p(x) = d(x) \cdot q(x) + r(x),$$$

where $$r$$ has a degree less than $$d$$. In this blog post, we restrict ourselves to division with a binomial, $$x - k$$, which means that $$r$$ is always a constant, and $$0$$ in the case where $$d$$ divides $$p$$.

One procedure for polynomial division is polynomial long division, which we can use to divide the polynomial $$p(x) = 2x^3 + 4x^2 + 11x + 3$$ with the binomial $$d(x) = x - 2$$, giving us the following result,3

$$$\begin{array}{ c c c c c c c c c } & & & & 2x^2 & + & 8x & + & 27 \\ \hline x - 2) & & 2x^3 & + & 4x^2 & + & 11x & + & 3 \\ & - & 2x^3 & + & 4x^2 & & & & \\ %\hline & & & & 8x^2 & + & 11x & & \\ & & & - & 8x^2 & + & 16x & & \\ %\hline & & & & & & 27x & + & 3 \\ & & & & & - & 27x & + & 54 \\ %\hline & & & & & & & & 57 \end{array}$$$

where we can read the quotient, $$2x^2 + 8x + 27$$, from the line above the numerator, $$2x^3 + 4x^2 + 11x + 3$$, and we can read the remainder, $$57$$, from the value at the bottom of the calculation. Lastly, we can verify the calculations by checking that the relation in Formula \ref{eq:poly-div-relation} is satisfied,

$\begin{equation*} 2x^3 + 4x^2 + 11x + 3 = (x - 2)(2x^2 + 8x + 27) + 57. \end{equation*}$

If we examine the intermediate results of the procedure, $$(2, 8, 27, 57)$$, i.e., the leftmost values of each step of the procedure, we can make out a similar recursive substitution scheme to what we saw in the case of polynomial evaluation,

\begin{align*} 2 &= 2\\ 8 &= 2 \cdot 2 + 4\\ 27 &= 8 \cdot 2 + 11\\ 57 &= 27 \cdot 2 + 3. \end{align*}

where each intermediate result is equal to the previous result multiplied by the second term of the denominator, $$x - 2$$, plus the next coefficient. This time, we assign the intermediate results on the left to the variable $$b_{i-1}$$, the last result to the variable $$r$$, the second term of the denominator to the variable $$k$$, and the coefficients of $$p$$ to the variable $$a_i$$, which yields the following set of equations,

\begin{align*} b_{2} &= a_3\\ b_{1} &= b_2 \cdot k + a_2\\ b_{0} &= b_1 \cdot k + a_1\\ r &= b_0 \cdot k + a_0. \end{align*}

These equations strongly suggest that we can divide $$p$$ with $$d$$ using the same recursive substitution procedure, as described in the evaluation case, spending just one addition and multiplication per term, which again reduces the number of operations to a minimum. Furthermore, we can put the substitution scheme above in a tabular format, similar to polynomial long division,

$$$\tag{7}\label{eq:horner-div-abstract} \begin{array}{ c | c c c c } & a_3 & a_2 & a_1 & a_0 \\ k & & b_2 \cdot k & b_1 \cdot k & b_0 \cdot k \\ \hline & a_3 & b_2 \cdot k + a_2 & b_1 \cdot k + a_1 & b_0 \cdot k + a_0 \\ & =b_2 & =b_1 & =b_0 & =r \end{array}$$$

where the coefficients of the polynomial are located at the top row, the second term of the denominator to the far left, and the coefficients of the resulting quotient, $$b_2,b_1,b_0$$, and the remainder, $$r$$, at the bottom row of the table.

We formalize the tabular representation in Formula \ref{eq:horner-div-abstract} as the following procedure,

which performs the exact same substitution scheme as in hornersPolyEvalAcc, except that it also aggregates the intermediate results and adds them to the result polynomial. Likewise, we define a wrapper function,

which sets the initial accumulator to the first coefficient and adds it to the result polynomial. Now, if we wanted to divide our initial polynomial $$p(x) = 2x^3 + 4x^2 + 11x + 3$$ with the binomial $$d(x) = x - 2$$, we would pass the list [2, 4, 11, 3] as the input polynomial cs and 2 as the input value x to hornersPolyDiv, from which we would get the result list [2, 8, 27, 57], where [2, 8, 27] are the coefficients of the quotient and 57 is the remainder. Thus, we have now defined Horner’s method for polynomial division as the procedures hornersPolyDivAcc and hornersPolyDiv.

### 4. Equivalence of the two Horner procedures

Due to the strong similarity between the procedure for polynomial evaluation and the procedure for polynomial division, we are interested in stating an equivalence relation between the two. As such, we note that the last element in the result polynomial of hornersPolyDiv is equal to the result of hornersPolyEval when given the same input,

Proving the relation requires us to first prove a similar equivalence relation between the underlying procedures hornersPolyEvalAcc and hornersPolyDivAcc, parameterized over the accumulator,

The equivalence can be proved by first proving the underlying theorem, using structural induction on the polynomial, cs', followed by case analysis on the polynomial, cs, in the original theorem.4 Incidentally, the above theorem also proves an implementation-specific version of the polynomial remainder theorem.

### 5. Conclusion

In this post, we have introduced Horner’s method for polynomial evaluation and polynomial division. Furthermore, we have also stated and proved an equivalence relation between the definition of Horner’s method for polynomial evaluation and polynomial division.

In our next post, we show how we can obtain Taylor polynomials using Horner’s method.

1. See “On Two Problems in Abstract Algebra Connected with Horner’s Rule” (1954) by Alexander Markowich Ostrowski and “Methods of computing values of polynomials” (1966) by Victor Ya Pan.

2. See “A new method of solving numerical equations of all orders, by continuous approximation” (1819) by William G. Horner.

3. Unfortunately, MathJax does not support partial horizontal lines (\cline) and thus we cannot format polynomial long division in the traditional way.

4. While we do not show each step of the proof in this post, a Coq implementation of Horner’s method and accompanying equivalence proof can be found in my Master’s thesis