Editing Multiplication algorithm (section)

==== General case with multiplication of N numbers ====

By exploring patterns after expansion, one see following:

<math display="block">\begin{alignat}{5} (x_1 B^{ m} + x_0) (y_1 B^{m} + y_0) (z_1 B^{ m} + z_0) (a_1 B^{ m} + a_0) &=
a_1 x_1 y_1 z_1 B^{4 m} &+ a_1 x_1 y_1 z_0 B^{3m} &+ a_1 x_1 y_0 z_1 B^{3 m} &+ a_1 x_0 y_1 z_1 B^{3 m} \\
&+ a_0 x_1 y_1 z_1 B^{3 m} &+ a_1 x_1 y_0 z_0 B^{2 m} &+ a_1 x_0 y_1 z_0 B^{2 m} &+ a_0 x_1 y_1 z_0 B^{2 m}\\ 
&+ a_1 x_0 y_0 z_1 B^{2 m} &+ a_0 x_1 y_0 z_1 B^{2 m} &+ a_0 x_0 y_1 z_1 B^{2 m} &+ a_1 x_0 y_0 z_0 B^{m\phantom{1}}\\
&+ a_0 x_1 y_0 z_0 B^{m\phantom{1}} &+ a_0 x_0 y_1 z_0 B^{m\phantom{1}} &+ a_0 x_0 y_0 z_1 B^{m\phantom{1}} &+ a_0 x_0 y_0 z_0 \phantom{B^{1 m}}
\end{alignat}</math>

Each summand is associated to a unique binary number from 0 to
<math> 2^{N+1}-1 </math>, for example <math> a_1 x_1 y_1 z_1 \longleftrightarrow 1111,\ a_1 x_0 y_1 z_0 \longleftrightarrow 1010 </math> etc. Furthermore; B is powered to number of 1, in this binary string, multiplied with m.

If we express this in fewer terms, we get:

<math display="block">\prod_{j=1}^N (x_{j,1} B^{ m} + x_{j,0})  = \sum_{i=1}^{2^{N+1}-1}\prod_{j=1}^N x_{j,c(i,j)}B^{m\sum_{j=1}^N c(i,j)} = \sum_{j=0}^{N}z_jB^{jm} 
</math>, where <math> c(i,j) </math> means digit in number i at position j. Notice that <math> c(i,j) \in \{0,1\} </math>

<math display="block">
\begin{align}
z_{0} &= \prod_{j=1}^N x_{j,0}
\\
z_{N} &= \prod_{j=1}^N x_{j,1}
\\
z_{N-1} &= \prod_{j=1}^N (x_{j,0} + x_{j,1}) -  \sum_{i \ne N-1}^{N} z_i
\end{align}
</math>