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Some textbook (Höhere Mathematik 1, Meyberg/Vachenauer, 6th edition) states that

the calculation of the functional values of a polynomial $f\colon \mathbb R\rightarrow\mathbb R$, $f(x) = a_0 + a_1x + a_2x^2 + \cdots + a_nx^n$, $n\ge1$, $a_i \in\mathbb R$, $a_n\ne 0$, according to Horner’s rule $$ f(x) = (\cdots((a_nx + a_{n-1})x + a_{n-2})x + \cdots + a_1)x + a_0 \tag{1}$$ includes the division by polynomials of form $x-b$

and refers to the following theorem:

Let $f(x) = a_nx^n + \cdots + a_1x + a_0$, $a_n\ne 0$, $n\ge 1$ and $b\in\mathbb R$. For numbers $c_n := a_n$, $c_{n-1} := c_nb + a_{n-1}$, $c_{n-2} := c_{n-1}b + a_{n-2}$, $\ldots$, $c_{n-k} := c_{n-k+1}b + a_{n-k}$, $\ldots$, $c_0 := c_1b + a_0$, there is

$$ c_0 = f(b) {\ \rm and\ } f(x) = (x-b)(c_nx^{n-1} + c_{n-1}x^{n-2} + \cdots + c_2x + c_1) + f(b).\tag{2}$$

I do not quite get the hang of the statement that “the calculation of the functional values by (1) includes the division of polynomials of form $x-b$”. I do recognize the term $x-b$ in (2) but I cannot relate it to (1). Can somebody give me some hint how to understand this?

Dennis Wydra
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  • My guess is that, if you look at Horner's rule above, and then you look what $c_n, c_{n-1}, c_{n-2}$ etc. are - you will notice that the pattern of calculating $c_i$'s closely follows the pattern of calculating $f(x)$ for $x=b$ using Horner's rule. For example, $c_n$ is the innermost $a_n$, $c_{n-1}$ is the innermost bracket $(a_nb+a_{n-1})$, $c_{n-2}$ will be the following outer bracket etc. - until you get that $c_0$ is the whole expression, i.e. $f(b)$. On the other hand, you can see $c_i$'s as the coefficients you obtain by performing division of $f(x)$ by $x-b$. –  Nov 29 '21 at 13:06
  • This is quite a roundabout and inconsequential argument IMHO, but there you go. A much simpler way to see it is just to divide $f(x)$ by $(x-b)$ to get a quotient $q(x)$ and a remainder $r$, which must be a constant, as $x-b$ is of degree one. Thus, $f(x)=(x-b)q(x)+r$. Substitute $x=b$ to conclude $f(b)=(b-b)q(b)+r=r$, so $r=f(b)$ and so $f(x)=(x-b)q(x)+f(b)$. –  Nov 29 '21 at 13:08

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