Editing Horner's method (section)

====Derivation====
In general, for a binary number with bit values (<math> d_3 d_2 d_1 d_0 </math>) the product is
<math display="block"> (d_3 2^3 + d_2 2^2 + d_1 2^1 + d_0 2^0)m = d_3 2^3 m + d_2 2^2 m + d_1 2^1 m + d_0 2^0 m. </math>
At this stage in the algorithm, it is required that terms with zero-valued coefficients are dropped, so that only binary coefficients equal to one are counted, thus the problem of multiplication or [[division by zero]] is not an issue, despite this implication in the factored equation:
<math display="block"> = d_0\left(m + 2 \frac{d_1}{d_0} \left(m + 2 \frac{d_2}{d_1} \left(m + 2 \frac{d_3}{d_2} (m)\right)\right)\right). </math>

The denominators all equal one (or the term is absent), so this reduces to
<math display="block"> = d_0(m + 2 {d_1} (m + 2 {d_2} (m + 2 {d_3} (m)))),</math>
or equivalently (as consistent with the "method" described above)
<math display="block"> = d_3(m + 2^{-1} {d_2} (m + 2^{-1}{d_1} (m + {d_0} (m)))). </math>

In binary (base-2) math, multiplication by a power of 2 is merely a [[arithmetic shift|register shift]] operation.  Thus, multiplying by 2 is calculated in base-2 by an [[arithmetic shift]].  The factor (2<sup>−1</sup>) is a right [[arithmetic shift]], a (0) results in no operation (since 2<sup>0</sup> = 1 is the multiplicative [[identity element]]), and a (2<sup>1</sup>) results in a left arithmetic shift.
The multiplication product can now be quickly calculated using only arithmetic shift operations, addition and [[subtraction]].

The method is particularly fast on processors supporting a single-instruction shift-and-addition-accumulate.  Compared to a C floating-point library, Horner's method sacrifices some accuracy, however it is nominally 13 times faster (16 times faster when the "[[canonical signed digit]]" (CSD) form is used) and uses only 20% of the code space.<ref>{{harvnb|Kripasagar|2008|p=62}}.</ref>