Editing Levinson recursion (section)

=== Using the backward vectors ===
The above steps give the ''N'' backward vectors for '''''M'''''. From there, a more arbitrary equation is:

: <math> \vec y = \mathbf M \  \vec x. </math>

The solution can be built in the same recursive way that the backwards vectors were built. Accordingly, <math>\vec x</math> must be generalized to a sequence of intermediates <math>\vec x^n</math>, such that <math>\vec x^N = \vec x</math>.

The solution is then built recursively by noticing that if

: <math> \mathbf T^{n-1}
  \begin{bmatrix}  x_1^{n-1}     \\
                   x_2^{n-1}     \\
                   \vdots   \\
                   x_{n-1}^{n-1} \\
  \end{bmatrix} =   
  \begin{bmatrix}  y_1     \\
                   y_2     \\
                   \vdots   \\
                   y_{n-1}
  \end{bmatrix}.</math>

Then, extending with a zero again, and defining an error constant where necessary:

: <math> \mathbf T^{n}
  \begin{bmatrix}  x_1^{n-1}     \\
                   x_2^{n-1}     \\
                   \vdots   \\
                   x_{n-1}^{n-1} \\
                   0
  \end{bmatrix} =   
  \begin{bmatrix}  y_1     \\
                   y_2     \\
                   \vdots   \\
                   y_{n-1} \\
                   \varepsilon_x^{n-1}
  \end{bmatrix}.</math>

We can then use the ''n''<sup>th</sup> backward vector to eliminate the error term and replace it with the desired formula as follows:

: <math> \mathbf T^{n} \left (
  \begin{bmatrix}  x_1^{n-1}     \\
                   x_2^{n-1}     \\
                   \vdots   \\
                   x_{n-1}^{n-1} \\
                   0 \\
  \end{bmatrix} + (y_n - \varepsilon_x^{n-1}) \  \vec b^n \right ) =   
  \begin{bmatrix}  y_1     \\
                   y_2     \\
                   \vdots   \\
                   y_{n-1} \\
                   y_n
  \end{bmatrix}.</math>

Extending this method until ''n'' = ''N'' yields the solution <math>\vec x</math>.

In practice, these steps are often done concurrently with the rest of the procedure, but they form a coherent unit and deserve to be treated as their own step.