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A method for obtaining the incomplete fraction and the remainder in the division of a polynomial
 
A method for obtaining the incomplete fraction and the remainder in the division of a polynomial
 +
\begin{equation}\label{eq:1}
 +
f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0
 +
\end{equation}
 +
by a binomial $x-a$, where all the coefficients $a,a_0,\ldots,a_n$ lie in a certain field, e.g. in the field of complex numbers. Any polynomial $f(x)$ can be uniquely represented in the form
 +
$$
 +
f(x) = g(x)(x-a) + r
 +
$$
 +
where $g(x) = b_{n-1}x^{n-1} + b_{n-2}x^{n-2} + \cdots + b_0$ is the incomplete fraction, while $r$ is the remainder which, according to the [[Bezout theorem]], equals $f(a)$. The coefficients of $g(x)$ and $r$ are calculated by the recurrence formulas
 +
\begin{equation}\label{eq:2}
 +
\left.{
 +
\begin{array}{c}
 +
b_{n-1} = a_n,\ \ b_{n-2} = a_{n-1} + a b_{n-1},\ \ldots\,,\ b_0 = a_1 + a b_1 \ ; \\
 +
r = a_0 + a b_0
 +
\end{array}
 +
}\right\rbrace\ \ .
 +
\end{equation}
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/h/h048/h048030/h0480301.png" /></td> <td valign="top" style="width:5%;text-align:right;">(1)</td></tr></table>
+
The following table
 
+
$$
by a binomial <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/h/h048/h048030/h0480302.png" />, where all the coefficients <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/h/h048/h048030/h0480303.png" /> lie in a certain field, e.g. in the field of complex numbers. Any polynomial <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/h/h048/h048030/h0480304.png" /> can be uniquely represented in the form
+
\begin{array}{|c|c|c|c|c|c|}
 
+
\hline \\
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/h/h048/h048030/h0480305.png" /></td> </tr></table>
+
a_n & a_{n-1} & \ldots & a_1 & a_0 \\
 
+
\hline \\
where <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/h/h048/h048030/h0480306.png" /> is the incomplete fraction, while <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/h/h048/h048030/h0480307.png" /> is the remainder which, according to the [[Bezout theorem|Bezout theorem]], equals <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/h/h048/h048030/h0480308.png" />. The coefficients of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/h/h048/h048030/h0480309.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/h/h048/h048030/h04803010.png" /> are calculated by the recurrence formulas
+
a & b_{n-1} & b_{n-2} & \ldots & b_0 \\
 
+
\hline
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/h/h048/h048030/h04803011.png" /></td> <td valign="top" style="width:5%;text-align:right;">(2)</td></tr></table>
+
\end{array}
 
+
$$
The following table:''''''<table border="0" cellpadding="0" cellspacing="0" style="background-color:black;"> <tr><td> <table border="0" cellspacing="1" cellpadding="4" style="background-color:black;"> <tbody> <tr> <td colname="1" style="background-color:white;" colspan="1"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/h/h048/h048030/h04803012.png" /></td> <td colname="2" style="background-color:white;" colspan="1"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/h/h048/h048030/h04803013.png" /></td> <td colname="3" style="background-color:white;" colspan="1"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/h/h048/h048030/h04803014.png" /></td> <td colname="4" style="background-color:white;" colspan="1"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/h/h048/h048030/h04803015.png" /></td> <td colname="5" style="background-color:white;" colspan="1"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/h/h048/h048030/h04803016.png" /></td> </tr> <tr> <td colname="1" style="background-color:white;" colspan="1"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/h/h048/h048030/h04803017.png" /></td> <td colname="2" style="background-color:white;" colspan="1"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/h/h048/h048030/h04803018.png" /></td> <td colname="3" style="background-color:white;" colspan="1"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/h/h048/h048030/h04803019.png" /></td> <td colname="4" style="background-color:white;" colspan="1"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/h/h048/h048030/h04803020.png" /></td> <td colname="5" style="background-color:white;" colspan="1"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/h/h048/h048030/h04803021.png" /></td> </tr> </tbody> </table>
+
whose upper line is given, while the lower line is filled in accordance with the formulas \eqref{eq:2}), is used in the computations. In fact, this method is identical with the method of Ch'in Chiu-Shao employed in medieval China. At the beginning of the 19th century it was rediscovered, almost simultaneously, by W.G. Horner [[#References|[1]]] and P. Ruffini [[#References|[1]]].
 
 
</td></tr> </table>
 
 
 
whose upper line is given, while the lower line is filled in accordance with the formulas (2), is used in the computations. In fact, this method is identical with the method of Ch'in Chiu-Shao employed in medieval China. At the beginning of the 19th century it was rediscovered, almost simultaneously, by W.G. Horner [[#References|[1]]] and P. Ruffini [[#References|[1]]].
 
  
 
====References====
 
====References====
<table><TR><TD valign="top">[1]</TD> <TD valign="top">  W.G. Horner,  ''Philos. Transact. Roy. Soc. London Ser. A'' , '''1'''  (1819)  pp. 308–335</TD></TR><TR><TD valign="top">[2]</TD> <TD valign="top">  P. Ruffini,  ''Mem. Coronata Soc. Ital. Sci.'' , '''9'''  (1802)  pp. 44–526</TD></TR></table>
+
<table>
 +
<TR><TD valign="top">[1]</TD> <TD valign="top">  W.G. Horner,  ''Philos. Transact. Roy. Soc. London Ser. A'' , '''1'''  (1819)  pp. 308–335</TD></TR>
 +
<TR><TD valign="top">[2]</TD> <TD valign="top">  P. Ruffini,  ''Mem. Coronata Soc. Ital. Sci.'' , '''9'''  (1802)  pp. 44–526</TD></TR>
 +
</table>
  
  
  
 
====Comments====
 
====Comments====
Horner's scheme is used for the efficient evaluation of polynomials. The straightforward evaluation of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/h/h048/h048030/h04803022.png" /> for a given value <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/h/h048/h048030/h04803023.png" /> by computing the powers <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/h/h048/h048030/h04803024.png" /> and forming the linear combination according to (1) would require <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/h/h048/h048030/h04803025.png" /> multiplications. However, by writing <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/h/h048/h048030/h04803026.png" /> in the  "nested"  form
+
Horner's scheme is used for the efficient evaluation of polynomials. The straightforward evaluation of $f(x)$ for a given value $x=a$ by computing the powers $a^2,a^3,\ldots,a^n$ and forming the linear combination according to \eqref{eq:1} would require $2n-1$ multiplications. However, by writing $f(x)$ in the  "nested"  form
 +
$$
 +
f(x) = x(\cdots(x(xa_n + a_{n-1}) + a_{n-2}) \cdots)+a_0\,,
 +
$$
 +
only $n$ (sequential) multiplications are involved. It is easily verified that using this nested representation the work consists of the computation steps given by \eqref{eq:2}. Since $r=f(a)$, the evaluation of $f(a)$ requires $n$ instead of $2n-1$ multiplications.
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/h/h048/h048030/h04803027.png" /></td> </tr></table>
+
====References====
 +
<table>
 +
<TR><TD valign="top">[a1]</TD> <TD valign="top">  F.B. Hildebrand,  "Introduction to numerical analysis" , McGraw-Hill  (1974)</TD></TR>
 +
<TR><TD valign="top">[a2]</TD> <TD valign="top">  F. Cajori,  "A history of mathematics" , Chelsea, reprint  (1980)</TD></TR>
 +
</table>
  
only <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/h/h048/h048030/h04803028.png" /> (sequential) multiplications are involved. It is easily verified that using this nested representation the work consists of the computation steps given by (2). Since <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/h/h048/h048030/h04803029.png" />, the evaluation of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/h/h048/h048030/h04803030.png" /> requires <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/h/h048/h048030/h04803031.png" /> instead of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/h/h048/h048030/h04803032.png" /> multiplications.
+
{{TEX|done}}
 
 
====References====
 
<table><TR><TD valign="top">[a1]</TD> <TD valign="top">  F.B. Hildebrand,  "Introduction to numerical analysis" , McGraw-Hill  (1974)</TD></TR><TR><TD valign="top">[a2]</TD> <TD valign="top">  F. Cajori,  "A history of mathematics" , Chelsea, reprint  (1980)</TD></TR></table>
 

Latest revision as of 18:09, 1 October 2017

A method for obtaining the incomplete fraction and the remainder in the division of a polynomial \begin{equation}\label{eq:1} f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \end{equation} by a binomial $x-a$, where all the coefficients $a,a_0,\ldots,a_n$ lie in a certain field, e.g. in the field of complex numbers. Any polynomial $f(x)$ can be uniquely represented in the form $$ f(x) = g(x)(x-a) + r $$ where $g(x) = b_{n-1}x^{n-1} + b_{n-2}x^{n-2} + \cdots + b_0$ is the incomplete fraction, while $r$ is the remainder which, according to the Bezout theorem, equals $f(a)$. The coefficients of $g(x)$ and $r$ are calculated by the recurrence formulas \begin{equation}\label{eq:2} \left.{ \begin{array}{c} b_{n-1} = a_n,\ \ b_{n-2} = a_{n-1} + a b_{n-1},\ \ldots\,,\ b_0 = a_1 + a b_1 \ ; \\ r = a_0 + a b_0 \end{array} }\right\rbrace\ \ . \end{equation}

The following table $$ \begin{array}{|c|c|c|c|c|c|} \hline \\ a_n & a_{n-1} & \ldots & a_1 & a_0 \\ \hline \\ a & b_{n-1} & b_{n-2} & \ldots & b_0 \\ \hline \end{array} $$ whose upper line is given, while the lower line is filled in accordance with the formulas \eqref{eq:2}), is used in the computations. In fact, this method is identical with the method of Ch'in Chiu-Shao employed in medieval China. At the beginning of the 19th century it was rediscovered, almost simultaneously, by W.G. Horner [1] and P. Ruffini [1].

References

[1] W.G. Horner, Philos. Transact. Roy. Soc. London Ser. A , 1 (1819) pp. 308–335
[2] P. Ruffini, Mem. Coronata Soc. Ital. Sci. , 9 (1802) pp. 44–526


Comments

Horner's scheme is used for the efficient evaluation of polynomials. The straightforward evaluation of $f(x)$ for a given value $x=a$ by computing the powers $a^2,a^3,\ldots,a^n$ and forming the linear combination according to \eqref{eq:1} would require $2n-1$ multiplications. However, by writing $f(x)$ in the "nested" form $$ f(x) = x(\cdots(x(xa_n + a_{n-1}) + a_{n-2}) \cdots)+a_0\,, $$ only $n$ (sequential) multiplications are involved. It is easily verified that using this nested representation the work consists of the computation steps given by \eqref{eq:2}. Since $r=f(a)$, the evaluation of $f(a)$ requires $n$ instead of $2n-1$ multiplications.

References

[a1] F.B. Hildebrand, "Introduction to numerical analysis" , McGraw-Hill (1974)
[a2] F. Cajori, "A history of mathematics" , Chelsea, reprint (1980)
How to Cite This Entry:
Horner scheme. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Horner_scheme&oldid=18660
This article was adapted from an original article by V.N. Remeslennikov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article